Maths Class X Question Bank

Question 1 of 38

1. Question
1 point(s)
In the figure, $\angle M=\angle N=46^o $. Express $x $ in terms of a, b and c, where a, b and c are lengths of $LM, MN $ and $NK $ respectively.
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Hint

In $\triangle LMK$ and $\triangle PNK$
$\angle LMK=\angle PNK$ (Corresponding angle)
$\angle LKM=\angle PKN$ (Common)
$\triangle LMK\cong\triangle PNK$ (AA similarly)
$\dfrac{ML}{NP}=\dfrac{MK}{NK}\\ \\ \dfrac{a}{x}=\dfrac{b+c}{c}\\ \\ x=\dfrac{ac}{b+c}$
Question 2 of 38

2. Question
1 point(s)
In the figure, LM || CB and LN || CD, prove that $\dfrac{AM}{AB}=\dfrac{AN}{AD} $.
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Hint

(1) In $\triangle ACB$
$LM||CB\\ \\ \dfrac{AL}{LC}=\dfrac{AM}{MB}$ …(1)
In $\triangle ACD$
$LN||CD\\ \\ \dfrac{AL}{LC}=\dfrac{AN}{ND}$ ….(2)
From (1) and (2)
$\dfrac{AM}{MB}=\dfrac{AN}{ND}\Rightarrow \dfrac{MB}{AM}=\dfrac{ND}{AN}\\ \\ \Rightarrow \dfrac{MB}{AM}+1=\dfrac{ND}{AN}+1\\ \\ \Rightarrow \dfrac{AB}{AM}=\dfrac{AD}{AN}\Rightarrow \dfrac{AM}{AB}=\dfrac{AN}{AD}$
Question 3 of 38

3. Question
1 point(s)
In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that :
(i) $\triangle ABC\sim\triangle AMP $
(ii) $\dfrac{CA}{PA}=\dfrac{BC}{MP} $.
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Hint

In $\triangle ABC$ and $\triangle AMP$
$\angle ABC=\angle AMP$ (each $90^o$ )
$\angle A=\angle A$ (common)
$\triangle ABC\sim\triangle AMP$ (AA similarly)
$\dfrac{CA}{PA}=\dfrac{BC}{MP}$ (Corresponding sides of similar triangles are proportional)
Question 4 of 38

4. Question
1 point(s)
In the figure, DE || OQ and DF || OR. Show that EF || QR.
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Hint

In $\triangle POQ$
$DE||OQ\\ \\ \Rightarrow \dfrac{PE}{EQ}=\dfrac{PD}{DO}$ …(1) (using BPT)
In $\triangle POR,\ DF||OR$
$\dfrac{PD}{DO}=\dfrac{PF}{FR}$ …(2) (using BPT)
comparing (1) and (2)
$\dfrac{PE}{EQ}=\dfrac{PF}{FR}\Rightarrow EF||QR$ (by converse of BPP)
Question 5 of 38

5. Question
1 point(s)
By using the converse of the basic proportionality theorem, show that the line joining mid-points of non-parallel sides of a trapezium is parallel to the parallel sides.
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Hint

In $\triangle ABF$ and $\triangle DGF$
$BF=DF$ ( $F$ is mid point of $BD$ )
$\angle ADF=\angle FDG$ (alt opp $\angle S$ )
$\triangle ABF\cong \triangle DGF$ (ASA)
$AF=FG$ and In $\triangle ACG$ as $AE=EC$ and $AF=FG$
i.e. $\dfrac{AE}{EC}=\dfrac{AF}{FG}$
Hence $EF||CD||AB$
Question 6 of 38

6. Question
1 point(s)
If three or more parallel lines are intersected by two transversals, the intercepts made by them on the transversal are proportional. Prove.
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Hint

In $\triangle AEF$
$CG||EF\\ \\ \Rightarrow \dfrac{AC}{CE}=\dfrac{AG}{GC}$
(using BPT) ….(1)
In $\triangle FBA,\ GD|| AB$
$\Rightarrow \dfrac{AG}{GF}=\dfrac{BD}{DF}$ …(2) (using BPT)
From (1) and (2)
$\dfrac{AC}{CE}=\dfrac{BD}{DF}$ (Proved)
Question 7 of 38

7. Question
1 point(s)
$ABC $ is a triangle in which $\angle A=90^o,\ AN\bot BC,\ AB=12\ cm $ and $AC=5\ cm $. Find the ratio of the area of $\triangle ANC $ and $\triangle ABC $.
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Hint

In $\triangle ANC$ and $\triangle ABC$
$\angle C=\angle C$ (common)
$\angle ANC=\angle BAC$ (each $90^o$ )
$\triangle ANC\sim\triangle BAC$ (AA similarly)
$\dfrac{ar\ (\triangle ANC)}{ar\ (\triangle BAC)}=\dfrac{AC^2}{BC^2}=\dfrac{5^2}{12^2}=\dfrac{25}{144}$
Question 8 of 38

8. Question
1 point(s)
ABCD is a square. F is the mid-point of AB.BE is the one-third of BC. If the area of the $\triangle $BFE is $108\ cm^2$, find the length of AC.
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Hint

Area of $\triangle FBE =108\ cm^2$
$\dfrac{1}{2}\times BE\times FB=108\\ \\ \dfrac{1}{2}\times\dfrac{x}{3}\times\dfrac{x}{2}=108\\ \\ x^2=108\times2\times3\times2\\ \\ x=\sqrt{1296}=36\ cm\\ \\ AC^2=x^2+x^2=2x^2=2\times(36)^2\\ \\ AC=36\sqrt2=36\times1.414\\ \\ =50904\ cm$
Question 9 of 38

9. Question
1 point(s)
In the given figure, $\angle B<90^o $ and segment $AD\bot BC $, show that $b^2=h^2+a^2+x^2-2ax $
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Hint

$AC^2=AD^2+DC^2\\ \\ b^2=h^2+(a-x)^2\\ \\ b^2=h^2+a^2-2ax+x^2\\ \\ b^2=h^2+a^2+x^2-2ax$
Question 10 of 38

10. Question
1 point(s)
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
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Hint

$AB=AC$ ( $\because\ \triangle$ is equilateral)
$\angle B=\angle C=60^o$ (common)
$\angle ADB=\angle ADC=90^o$
$\triangle ABD\cong\triangle ACD$ (AAS)
$BD=DC=\dfrac{1}{2}BC$
In $\triangle ABD$
$AB^2=AD^2+BD^2=AD^2+\left(\dfrac{BC}{2}\right)^2\\ \\ AB^2=\dfrac{4AD^2+BC^2}{4}\\ \\ 4AB^2=4AD^2+BC^2\Rightarrow 4AB^2-BC^2=4AD^2\\ \\ \Rightarrow 3AB^2=4AD^2$
Question 11 of 38

11. Question
1 point(s)
Using BPT, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
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Hint

$PQ||BC$
$\dfrac{AQ}{QC}=\dfrac{AP}{PB}$ (using BPT)
$\dfrac{AQ}{QC}=\dfrac{1}{1}\ (\because\ AP=PB)\\ \\ \Rightarrow AQ=QC$
or $Q$ is the mid point of $AC$
Question 12 of 38

12. Question
1 point(s)
If $\triangle ABC\sim\triangle DEF $ and, $AL $ and $DM $ are their corresponding medians, then show that $\dfrac{AL}{DM}=\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF} $.
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Hint

$\angle A=\angle D,\ \angle B=\angle E,\ \angle C=\angle F$ (given)
$\triangle ABC\sim\triangle DEF$ (AA)
$\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF}$ ….(1)
In $\triangle ALB$ and $\triangle DME$
$\angle ALB=\angle DME=90^o$
$\angle B=\angle E$ (given)
$\triangle ALB\sim\triangle DME$ (AA)
$\dfrac{AB}{DE}=\dfrac{AL}{DM}$ ….(2)
From (1) and (2)
$\dfrac{AL}{DM}=\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF}$
Question 13 of 38

13. Question
1 point(s)
If $\triangle ABC\sim\ \triangle DEF $, then show that $\dfrac{perimeter\ of\ \triangle ABC}{perimeter\ of\ \triangle DEF}=\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF} $.
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Hint

$\triangle ABC\sim\triangle DEF\\ \\ \dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{CA}{FD}=\dfrac{m}{n}$
$AB=DE\left(\dfrac{m}{n}\right)$ ….(1)
$BC=EF\left(\dfrac{m}{n}\right)$ …(2)
$CA=FO\left(\dfrac{m}{n}\right)$ ….(3)
$AB+BC+CA=(DE+EF+FD)\dfrac{m}{n}$
$\dfrac{Perimeter\ of\ ABC}{Perimeter\ of\ DEF}=\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF}$
Question 14 of 38

14. Question
1 point(s)
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 3CD, find the ratio of the areas of triangles AOB and COD.
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Hint

In $\triangle AOB$ and $\triangle COD$
$\angle AOB=\angle COD\\ \\ \angle OAB=\angle OCD$
$\triangle AOB\sim\triangle COD$ (AA)
$\dfrac{ar\ (\triangle AOB)}{ar\ (\triangle COD)}=\left(\dfrac{2CD}{CD}\right)^2=\dfrac{4}{1}$
Question 15 of 38

15. Question
1 point(s)
In $\triangle $ABC, in figure, PQ meets AB in P and AC in Q. If If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of $\triangle $APQ is one sixteenth of the area of $\triangle $ABC.
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Hint

In $\triangle APQ$ and $\triangle ABC$
$\angle A=\angle A$ (common)
$\dfrac{AP}{AB}=\dfrac{AQ}{AC}$
$\triangle APQ\sim\triangle ABC$ (SAS)
$\dfrac{ar\ (\triangle APQ)}{ar\ (\triangle ABC)}=\dfrac{1^2}{4^2}=\dfrac{1}{16}\\ \\ ar(\triangle APQ)=\dfrac{1}{16}\times ar\ (\triangle ABC)$
Question 16 of 38

16. Question
1 point(s)
In the figure, PA, QB and RC are perpendiculars to AC. Prove that $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{z} $
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Hint

Let $AB=a,\ BC=b$
In $\triangle CQB$ and $\triangle CPA$
$\angle QCB=\angle PCA$ (common)
$\angle QBC=\angle PAC$ (each $90^o$ )
$\triangle CQB\sim\triangle CPA$ (AA)
$\dfrac{QB}{PA}=\dfrac{CB}{CA}$
$\Rightarrow \dfrac{y}{x}=\dfrac{b}{a+b}$ …(1)
Similarly $\triangle AQB\sim\triangle ARC$ (AA)
$\dfrac{QB}{RC}=\dfrac{AB}{AC}$
$\Rightarrow \dfrac{y}{z}=\dfrac{a}{a+b}$ …(2)
Adding eqn (1) and (2)
$\dfrac{y}{x}+\dfrac{y}{z}=\dfrac{a}{a+b}+\dfrac{b}{a+b}=\dfrac{b+a}{a+b}\\ \\ y\left(\dfrac{1}{x}+\dfrac{1}{z}\right)=1\Rightarrow \dfrac{1}{x}+\dfrac{1}{z}=\dfrac{1}{y}$
Question 17 of 38

17. Question
1 point(s)
In an equilateral triangle $ABC, D$ is a point on side $BC $ such that $3BD = BC$. Prove that $9AD^2=7AB^2$.
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Hint

$BD=\dfrac{x}{3}$
In $\triangle AEB$ and $\triangle AEC$
$AE=AE$ (common)
$AB=AC\\ \\ \angle AEB=\angle AEC$
$\triangle AEB\cong\triangle AEC$ (RHS)
$BE=EC$ (cpct)
$BE=EC=\dfrac{x}{2},\ BD+DE=\dfrac{x}{2}\\ \\ \dfrac{x}{3}+DE=\dfrac{x}{2}\Rightarrow DE=\dfrac{x}{6} \\ \\ AB^2=AE^2+BE^2\\ \\ x^2=AE^2+\dfrac{x^2}{4}\Rightarrow AE^2=\dfrac{3x^2}{4}$
In $\triangle AED$
$AD^2=\dfrac{3x^2}{4}+\dfrac{x^2}{36}\Rightarrow AD^2=\dfrac{7x^2}{9}\Rightarrow 9AD^2=\dfrac{7x^2}{9}\Rightarrow 9AD^2=7AB^2$
Question 18 of 38

18. Question
1 point(s)
In $\triangle $PQR,PD $\bot $ QR such that D lies on QR. If PQ = a, PR = b, QD = C and DR = d and a, b, c, d are positive units, prove that (a+b) (a-b) = (c+d) (c-d).
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Hint

In $\triangle PQD$
$PQ^2=PD^2+QD^2$
$\Rightarrow PD^2=a^2-c^2$ ….(1)
In $\triangle PDR$
$PD^2=PR^2-DR^2=b^2-d^2$ …(2)
From (1) and (2)
$a^2-c^2=b^2-d^2\\ a^2-b^2=c^2-d^2$
Question 19 of 38

19. Question
1 point(s)
In figure, $BL $ and $CM $ are medians of $\triangle ABC $. right angled at $A $. Prove that $4(BL^2+CM^2) = 5BC^2$
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Hint

In $\triangle ABL$
$BL^2=AB^2+AL^2\\ \\ =AB^2+\left(\dfrac{AC}{2}\right)^2$
In $\triangle ACM$
$CM^2=AC^2+AM^2=AC^2+\left(\dfrac{AB}{2}\right)^2\\ \\ BL^2+CM^2=AB^2+AC^2+\dfrac{AC^2}{4}+\dfrac{AB^2}{4}\\ \\ \Rightarrow 4(BL^2+CM^2)=5(AB^2+AC^2)=5BC^2$
Question 20 of 38

20. Question
1 point(s)
In the figure, ABC is an isosceles triangle right angled at B. Two equilateral triangles are constructed with side BC and AC. Prove that : $\mathrm{ar\ \triangle BCD=\dfrac{1}{2}ar\ \triangle ACE} $
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Hint

$AB=BC\\ \\ AB^2-BC^2\\ \\ AC^2=AB^2+BC^2\\ \\ AC^2=2AB^2=2BC^2\\ \\ \dfrac{AC^2}{BC^2}=\dfrac{2}{1}$ …(1)
$\dfrac{ar\ (\triangle ACE)}{ar\ (\triangle BCD)}=\dfrac{\frac{\sqrt3}{4}a^2}{\frac{\sqrt3}{4}a^2}=\dfrac{\frac{\sqrt3}{4}AC^2}{\frac{\sqrt3}{4}BC^2}=\dfrac{2}{1}\\ \\ (ar\ (\triangle BCD)=\dfrac{1}{2}\ ar\ (\triangle ACE))$
Question 21 of 38

21. Question
1 point(s)
In figure, ABC is right triangle right angled at C. Let BC = a, CA = b, AB = c and let p be the length of perpendicular from C on AB. Prove that
(i) $cp=ab $
(ii) $\dfrac{1}{p^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2} $
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Hint

Area of $\triangle ABC=\dfrac{1}{2}\times AB\times CD=\dfrac{1}{2}\times C\times P=\dfrac{1}{2}CP$
Area of $ABC=\dfrac{1}{2}\times BC\times AC=\dfrac{1}{2}\times a\times b=\dfrac{ab}{2}$
So, $\dfrac{CP}{2}=\dfrac{ab}{2}\Rightarrow CP=ab$
$AB^2=BC^2+AC^2\Rightarrow C^2=a^2+b^2\\ \\ \Rightarrow \left(\dfrac{ab}{p}\right)^2=a^2+b^2\\ \\ \Rightarrow \dfrac{a^2b^2}{p^2}=a^2+b^2\Rightarrow \dfrac{1}{p^2}=\dfrac{1}{b^2}+\dfrac{1}{a^2}$
Question 22 of 38

22. Question
1 point(s)
In the figure, PQR is a triangle in which $QM\bot PR $ and $PR^2-PQ^2=QR^2 $, prove that $QM^2=PM\times MR $.
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Hint

$PR^2-PQ^2=QR^2$ ….(1)
In $\triangle QMR$
$QR^2=QM^2+MR^2$ …(2)
From (1) and (2)
$PR^2-PQ^2=QM^2+MR^2\\ \\ QM^2=PR^2-PQ^2-MR^2\\ \\ QM^2=(PM+MR)^2-PQ^2-MR^2\\ \\ QM^2=PM^2+MR^2+2PM\cdot MR-PQ^2-MR^2\\ \\ 2QM^2=2PM\cdot MR$
Question 23 of 38

23. Question
1 point(s)
In figure, DE || BC and AD : DB = 5 : 4. Find $\dfrac{ar(\triangle DFE)}{ar(\triangle CFB)} $.
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Hint

$\triangle ADE\sim\triangle ABC\\ \\ \dfrac{AD}{AB}=\dfrac{DE}{BC}\Rightarrow \dfrac{5x}{9x}=\dfrac{DE}{BC}\\ \\ \Rightarrow \dfrac{DE}{DC}=\dfrac{5}{9}$ …(1)
In $\triangle DEF$ and $\triangle BFC$
$\angle EDF=\angle FCB$ (alt. $\angle S$ )
$\angle DFE=\angle BFC$ (VOA)
$\triangle DEF\sim\triangle BFC\\ \\ \dfrac{ar\ (\triangle DEF)}{ar\ (\triangle BFC)}=\left(\dfrac{DC}{BC}\right)^2=\left(\dfrac{5}{9}\right)^2=\dfrac{25}{81}$
Question 24 of 38

24. Question
1 point(s)
In the figure, $PQR $ is a right angled triangle in which $Q=90^o $. If $QS=SR $, show that $PR^2-4PS^2=3PQ^2 $.
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Hint

$QS=SR\\ \\ PR^2=PQ^2+QR^2\\ \\ PS^2=PQ^2+QS^2=PQ^2+\left(\dfrac{QR}{2}\right)^2\\ \\ PS^2=PQ^2+\dfrac{QR^2}{4}\\ \\ 4PS^2=4PQ^2+QR^2\\ \\ QR^2=4PS^2-4PQ^2\\ \\ PR^2=PQ^2+(4PS^2-4PQ^2)\\ \\ PR^2=4PS^2-3PQ^2$
Question 25 of 38

25. Question
1 point(s)
In the figure, l || m and line segments AB, CD and EF are concurrent at P. Prove that : $\dfrac{AE}{BF}=\dfrac{AC}{BD}=\dfrac{CE}{FD} $.
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Hint

In $\triangle APC$ and $\triangle BPD$
$\angle APC=\angle BPD$ (VOA)
$\angle PAC=\angle PBD$ (alt $\angle S$ )
$\triangle APC\sim\triangle BPD$ (AAA)
$\dfrac{AP}{PB}=\dfrac{AC}{BD}=\dfrac{PC}{AD}$ …(1)
In $\triangle APE$ and $\triangle BPF$
$\angle APE=\angle BPF$ (VOA)
$\angle PAE=\angle PBF$ (alt. $\angle S$ )
$\triangle APE\sim\triangle BPF$ (AA)
$\dfrac{AP}{PB}=\dfrac{AE}{BF}=\dfrac{PE}{PF}$ ….(2)
Similarly $\dfrac{PE}{PF}=\dfrac{PC}{PD}=\dfrac{EC}{FD}$ ….(3)
From (1), (2), (3)
$\dfrac{AE}{BF}=\dfrac{AC}{BD}=\dfrac{CE}{FD}$
Question 26 of 38

26. Question
1 point(s)
In the figure, $\angle QPR=90^o,\angle PMR=90^o,\ QR=26\ cm,\ PM=8\ cm,\ MR=6\ cm $. Find area $(\triangle PQR) $.
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Hint

In $\triangle PMR$
$PR=\sqrt{6^2+8^2}=\sqrt{100}=10\\ \\ QR^2=PQ^2+PR^2\Rightarrow PQ^2=676-100=576\\ \\ PQ=24$
Area of $\triangle PQR=\dfrac{1}{2}\times24\times10=120\ cm^2$
Question 27 of 38

27. Question
1 point(s)
In $\triangle $ABC,D and E are two points lying on side AB such that AD = BE. If DP||BC and EQ || AC, then prove that PQ||AB.
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Hint

In $\triangle ABC$ ,
$DP||BC$ and $EQ||AC\\ \\ \dfrac{AD}{DB}=\dfrac{AP}{PC}$ and $\dfrac{BE}{EA}=\dfrac{BQ}{QC}$
$\Rightarrow \dfrac{AD}{DB}=\dfrac{AP}{PC}$ and $\dfrac{AD}{DB}=\dfrac{BQ}{QC}\\ \\ \Rightarrow \dfrac{AP}{PC}=\dfrac{BQ}{QC}$
So by the converse of BPT $PQ||AB$
Question 28 of 38

28. Question
1 point(s)
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
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Hint

$\angle AOB=\angle BOC=\angle COD=\\ \\ \angle DOA=90\\ \\ OA=\dfrac{1}{2}AC$ and $OB=\dfrac{1}{2}BD$
In $\triangle AOB$
$AB^2=AO^2+OB^2=\left(\dfrac{1}{2}AC\right)^2+\left(\dfrac{1}{2}BD\right)^2\\ \\ \Rightarrow AB^2=\dfrac{1}{4}(AC^2+BD^2)\\ \\ \Rightarrow 4AB^2=AC^2+BD^2\\ \\ \Rightarrow AB^2+BC^2+CD^2+DA^2=AC^2+BD^2$
Question 29 of 38

29. Question
1 point(s)
In the figure, $DEFG $ is a square and $\angle BAC=90^o $. Show that $DE^2=BD\times EC $.
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Hint

In $\triangle AFG$ and $\triangle DBG$
$\angle GAF=\angle BDG\ (90)$
$\angle AGF=\angle DBG$ (Corresponding angles)
$\triangle AFG\sim\triangle DBG$ (AA) …(1)
In $\triangle AGF$ and $\triangle EFC$
$\angle AFG=\angle CEF\ (90)$
$\angle AFG=\angle ECF$ (corresponding angles)
$\triangle AGF\sim\triangle EFC$ (AA)….(2)
From (1) and (2)
$\triangle DBG\sim\triangle EFC\\ \\ \dfrac{BD}{EF}=\dfrac{DG}{EC}\Rightarrow \dfrac{BD}{DE}=\dfrac{DE}{EC}$ (DEFG is a square)
$DE^2=BD\times EC$
Question 30 of 38

30. Question
1 point(s)
In the figure, $\triangle PQR $ is right angled at $Q $ and the points $S $ and $T $ trisect the side $QR $. Prove that $8PT^2=3PR^2 + 5PS^2 $.
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Hint

Let $QS=TS=RT=x$
$QR=3x$ and $QT=2x$
In $\triangle PQR$
$PR^2=PQ^2+QR^2\Rightarrow PR^2=PQ^2+9x^2$
In $\triangle PQS$
$PS^2=PQ^2+QS^2=PQ^2+x^2$
In $\triangle PQT,\ PT^2=PQ^2+OT^2=PQ^2+4x^2\\ \\ 8PT^2=8(PQ^2+4x^2)=8PQ^2+32x^2\\ \\ 3PR^2+5PS^2=3(PQ^2+9x^2)+5(PQ^2+x^2)\\ \\ =8PQ^2+32x^2$
Question 31 of 38

31. Question
1 point(s)
In the given figure, O is a point in the interior of a triangle ABC, OD $\bot $ BC,OE $\bot $ AC and OF $\bot $ AB. Show that
(i) $OA^2+OB^2+OC^2-OD^2-OE^2-OF^2= AF^2+ BD^2+CE^2 $
(ii) $AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2 $
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Hint

$OA^2=AF^2+OF^2$ …(1)
$OB^2=OD^2=OD^2+BD^2$ …(2)
$OC^2=OE^2+EC^2$ …(3)
Adding (1), (2), (3) we get
$(OA)^2+(OB)^2+(OC)^2=AF^2+OF^2+OD^2+BD^2+OE^2+EC^2\\ \\ OA^2+OB^2+OC^2-OD^2-OE^2-OF^2=AF^2+BD^2+CE^2$ (Proved)
(ii) From (1) we have
$AF^2+BD^2+CE^2=OA^2+OB^2+OC^2-OD^2-OE^2-OF^2\\ \\ =(OA^2-OE^2)+(OB^2-OF^2)+(OC^2-OD^2)\\ \\ =AE^2+BF^2+CD^2$ (Proved)
Question 32 of 38

32. Question
1 point(s)
In a right triangle ABC, right angled at C. P and Q are points on the sides CA and CB respectively which divide these sides in the ratio 1 : 2. Prove that :
(i) $ 9AQ^2 = 9AC^2 + 4BC^2 $
(ii) $9BP^2 = 9BC^2 + 4AC^2 $
(iii) $9(AQ^2 + BP^2) = 13AB^2 $
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Hint

(i) $CQ=\dfrac{2}{3}BC,\ PC=\dfrac{2}{3}AC$
$AQ^2=AC^2+CQ^2\\ \\ =AC^2+\left(\dfrac{2}{3}BC\right)^2$
$AQ^2=AC^2+\dfrac{4}{9}BC^2\\ \\ 9AQ^2=9\ AC^2+4BC^2$ (Proved) ….(1)
(ii) In $\triangle PCB,\ BP^2=BC^2+PC^2=BC^2+\left(\dfrac{2}{3}AC\right)^2$
$BP^2=BC^2+\dfrac{4}{9}AC^2\Rightarrow 9BP^2=9BC^2+4AC^2$ ….(2)
(ii) Adding (1) and (2)
$9AQ^2+9BP^2=9AC^2+9BC^2+4BC^2+4AC^2\\ \\ 9(AQ^2+BP^2)=13(AC^2+BC^2)\\ \\ 9(AQ^2+BP^2)=13\ AB^2$
Question 33 of 38

33. Question
1 point(s)
In the given figure, in $\triangle $PQR XY || QR. PX = 1 cm, XQ = 3 cm, YR = 4.5 cm and QR = 9 cm. Find PY and XY. Further if the area of $\triangle $PXY is ‘A’ $cm^2 $, find in terms of A the area of $\triangle $PQR and the area of trapezium XYRQ.
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Hint

In $\triangle Pxy$ and $\triangle PQR$
$\angle Pxy=\angle PQR\\ \angle Pyx=\angle PRQ$ (corresponding angles)
$\triangle Pxy\sim\triangle PQR$ (AA)
$\dfrac{Px}{PQ}=\dfrac{xy}{QR}\Rightarrow \dfrac{1}{4}=\dfrac{xy}{QR}\Rightarrow xy=\dfrac{9}{4}=2.25\ cm$
$\dfrac{ar\ (\triangle Pxy)}{ar\ (\triangle PQR)}=\left(\dfrac{Px}{PQ}\right)^2=\left(\dfrac{1}{4}\right)^2=\dfrac{1}{16}\\ \\ ar\ \triangle PQR=16\ cm^2\\ \\ ar\ (trapezium\ XQRY)=ar(\triangle PQR)-ar(\triangle PXY)\\ \\ =(16A-A)\ cm^2\\ \\ =15A\ cm^2$
Question 34 of 38

34. Question
1 point(s)
If A be the area of a right triangle and a one of the sides containing the right angle, prove that the length of the altitude on the hypotenuse is $\dfrac{2Aa}{\sqrt{a^4+4A^2}} $.
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Hint

Base of the right angled $\triangle =a$ unit
$A=\dfrac{1}{2}\times a\times b\\ \\ \Rightarrow h=\dfrac{2A}{a}\\ \\ (Hypotenuse)^2=a^2+\left(\dfrac{2A}{a}\right)^2\\ \\ a^2+\dfrac{4A^2}{a^2}\\ \\ Hypotenuse=\sqrt{\dfrac{a^4+4A^2}{a^2}}=\dfrac{1}{a}\sqrt{a^4+4A^2}$
Area $=\dfrac{1}{2}\times\dfrac{1}{a}\sqrt{a^4+4A^2}$
$2AQ=\sqrt{a^4+4A^2}\times altitude$
Altitude $=\dfrac{2Aa}{\sqrt{a^4+4A^2}}$
Question 35 of 38

35. Question
1 point(s)
Prove that if a line is drawn parallel to one side of a triangle to intersect the other sides in distinct points, the other two sides are divided in the same ratio.
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Hint

Area of $\triangle =\dfrac{1}{2}\times base\times height$
area of $\triangle P\times Y=\dfrac{1}{2}\times P\times YN$ ….(1)
Also area of $\triangle P\times Y=\dfrac{1}{2}\times PY\times XN$ …(2)
Similarly $ar(\triangle Q\times Y)=\dfrac{1}{2}\times Q\times XNY$ …(3)
$ar\ (\triangle R\times Y)=\dfrac{1}{2}\times YR\times XM$ ….(4)
Dividing (1) by (3) we get
$\dfrac{ar(\triangle P\times y)}{ar\ (\triangle Q\times y)}=\dfrac{\frac{1}{2}\times P\times XYN}{\frac{1}{2}\times Q\times XYN}=\dfrac{PX}{QX}$ …(5)
Again dividing (2) by (4) we get
$\dfrac{ar(\triangle PXY)}{ar\ (\triangle RXY)}=\dfrac{PY}{YR}$ ….(6)
$ar\ (\triangle QXY)=ar\ (\triangle RXY)$ …(7)
From (5), (6), (7) we get $\dfrac{PX}{XQ}=\dfrac{PY}{YR}$ (Proved)
Question 36 of 38

36. Question
1 point(s)
State and prove converse of Pythagoras theorem.
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Hint

Area of $\triangle =\dfrac{1}{2}\times base\times height$
area of $\triangle P\times Y=\dfrac{1}{2}\times P\times YN$ ….(1)
Also area of $\triangle P\times Y=\dfrac{1}{2}\times PY\times XN$ …(2)
Similarly $ar(\triangle Q\times Y)=\dfrac{1}{2}\times Q\times XNY$ …(3)
$ar\ (\triangle R\times Y)=\dfrac{1}{2}\times YR\times XM$ ….(4)
Dividing (1) by (3) we get
$\dfrac{ar(\triangle P\times y)}{ar\ (\triangle Q\times y)}=\dfrac{\frac{1}{2}\times P\times XYN}{\frac{1}{2}\times Q\times XYN}=\dfrac{PX}{QX}$ …(5)
Again dividing (2) by (4) we get
$\dfrac{ar(\triangle PXY)}{ar\ (\triangle RXY)}=\dfrac{PY}{YR}$ ….(6)
$ar\ (\triangle QXY)=ar\ (\triangle RXY)$ …(7)
From (5), (6), (7) we get $\dfrac{PX}{XQ}=\dfrac{PY}{YR}$ (Proved)
$DE||BC$
By BPT
$\dfrac{AD}{DB}=\dfrac{AE}{EC}\Rightarrow \dfrac{AD}{DB}=\dfrac{AE}{DB}\ (\because\ BD=CE)\\ \\ \Rightarrow AD=AE$
Adding $DB$ on both the sides
$AD+DB=AE+DB\\ \\ \Rightarrow AD+DB=AE+EC\ (\because\ BD=CE)\\ \\ \Rightarrow AD=AC$
$\triangle ABC$ is isosceles $\triangle$
Question 37 of 38

37. Question
1 point(s)
Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Using the above result do the following :
In the figure, DE || BC and BD = CE.
Prove that $\triangle $ABC is an isosceles triangle.
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Hint

In $\triangle BMC$ and $\triangle EMD$
$MC=MD$
$\angle MBC=\angle MED$ (alt $\angle S$ )
$\triangle BMC\cong \triangle EMD$ (AAS)
$BC=DE$ (cpct)
$BC=AD$ ( $\because ABCD$ is a parallelogram)
In $\triangle AEL$ and $\triangle CBL$
$\angle ALE=\angle CLB$ (VOA)
$\angle EAL=\angle BCL$ (alt $\angle S$ )
$\triangle AEL\sim\triangle CBL$ (AA)
$\dfrac{EL}{BL}=\dfrac{AE}{CB}\Rightarrow \dfrac{EL}{DL}=\dfrac{2BC}{BC}\Rightarrow EL=2BL$
Question 38 of 38

38. Question
1 point(s)
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD produced in E. Prove that EL = 2BL.
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Hint

In a triangle if the square of one side is equal to the sum of the square of the other two sides then the angle opposite to the first side is a right angle
To prove $\angle Q=90^o$
Proof
$XZ^2=XY^2+YZ^2\\ \\ \Rightarrow XZ^2=PQ^2+QR^2$ …(1) ( $\because$ Since $XY=PQ$ and $YZ=QR$ )
But $PR^2=PQ^2+QR^2$ (given) …(2)
From (1) and (2)
$PR^2=XZ^2\Rightarrow PR=XZ$
In $\triangle PQR$ and $\triangle XYZ$
$PQ=XY\\ \\ QR=YZ\\ \\ PR=XZ\\ \\ \triangle PQR\cong \triangle XYZ\\ \\ \angle Q=\angle Y=90^o$