Maths Class X Question Bank

Question 1 of 38

1. Question
1 point(s)
$D $ is a point of the side $BC $ of a triangle ABC such that $\angle ADC=\angle BAC $. Show that $CA^2=CB.CD. $
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Hint

In $\triangle BAC$ and $\triangle ADC$
$\angle ACB=\angle ACD$ (common angle)
$\angle BAC=\angle ADC$ (given)
$\triangle BAC\sim\triangle ADC$
$\dfrac{BA}{AD}=\dfrac{AC}{CD}=\dfrac{BC}{AC}$ (Corresponding sides are proportional)
$\dfrac{AC}{CD}=\dfrac{BC}{AC}\Rightarrow AC^2=BC\ CD$
Question 2 of 38

2. Question
1 point(s)
$S $ and $ T$ are points on sides $PR $ and $QR $ of $\triangle PQR $ such that $\angle P=\angle RTS $. Show that $\triangle RPQ\sim\triangle RTS $.
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Hint

In $\triangle RPQ$ and $\triangle RTS$
$\angle P=\angle RTS$ (given)
$\angle PRQ=\angle TRS=\angle R$ (given)
$\triangle RPQ\sim\triangle RTS$ (AA similarly)
Question 3 of 38

3. Question
1 point(s)
In the given figure, AB||QR. Find the length of PB.
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Hint

$AB||QR\\ \\ \angle PAB=\angle PQR\\ \\ \angle PBA=\angle PRQ$
$\triangle PAB\sim\triangle PQR$ (AA Similarly)
$\dfrac{AB}{QR}=\dfrac{PB}{PR}\\ \\ \dfrac{3}{9}=\dfrac{PB}{6}\Rightarrow PB=2\ cm$
Question 4 of 38

4. Question
1 point(s)
In the given figure, if AB||CD, find the value of x.
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Hint

In $\triangle AOB$ and $\triangle DOC$
$\angle BAO=\angle DCO$ (alternate angle)
$\angle ABO=\angle CDO$ (alternate angle)
$\triangle AOB\sim\triangle COD$ (AA)
$\dfrac{AO}{OC}=\dfrac{OB}{OD}\Rightarrow \dfrac{4}{4x-2}=\dfrac{x+1}{2x+4}\\ \\ \Rightarrow 4x^2-6x-18=0\\ \\ \Rightarrow x=\dfrac{-3}{2},3$
Question 5 of 38

5. Question
1 point(s)
In the given figure, $\angle ABC=90^o $ and $BD\bot AC $. If $BD=8\ cm $ and $AD=4\ cm $, find $CD $.
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Hint

$\angle BDC=90^o,\ \angle ABC=90^o$
Let $\angle BCD=x$
$\angle DBC=90-x$
Also $\angle ABD=x$
$\angle BAD=90-x$
In $\triangle BDC$ and $\triangle ADB$
$\angle BDC-\angle BDA$ (each $90^o$ )
$\angle DBC=\angle DAB$ (each 90-x)
$\triangle BDC\sim\triangle ADB$ (AA)
$\dfrac{BD}{CD}=\dfrac{AD}{BD}\\ \\ \dfrac{8}{CD}=\dfrac{4}{8}\Rightarrow CD=16\ cm$
Question 6 of 38

6. Question
1 point(s)
In a right angled triangle with sides a and b and hypotenuse c, the altitude drawn on the hypotenuse is x. Prove that ab=cx.
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Hint

In $\triangle ACB$ and $\triangle CDB$
$\angle B=\angle B$ (common)
$\angle ACB=\angle CDB$ (each 90)
$\triangle ACB\sim\triangle CDB$ (AA Similarly)
$\dfrac{AB}{CD}=\dfrac{AC}{CB}\\ \\ \dfrac{a}{x}=\dfrac{c}{b}\\ \\ \Rightarrow ab=cx$
Question 7 of 38

7. Question
1 point(s)
In the given figure, $\dfrac{AD}{DB}=\dfrac{AE}{EC} $ and $\angle ADE=\angle ACB $. Prove that $\triangle ABC $ is an isosceles triangle.
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Hint

$\dfrac{AD}{DB}=\dfrac{AE}{EC}$
$DE||BC$ (converse of BPT)
$\angle ADE=\angle ABC$ (corresponding angles) ….(1)
$\angle ADB=\angle ABC=\angle ACB$ …(2)
From (1) and (2)
$\angle ABC=\angle ACB$
$AB=AC$ (opposite sides of opposite angle are equal)
Question 8 of 38

8. Question
1 point(s)
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\mathrm{\triangle ABE\sim\triangle CFB} $.
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Hint

$\angle A=\angle C$ (opposite angles of a parallelogram are equal) …(1)
$AD||BC$
$\angle AEB=\angle CBF$ (alternate angles) …(2)
In $\triangle ABE$ and $\triangle CFB$
$\angle A=\angle C$ (From (1))
$\angle AEB=\angle CBF$ (From (2))
$\triangle ABE\sim\triangle CFB$
Question 9 of 38

9. Question
1 point(s)
P and Q are points on sides AB and AC respectively of $\triangle $ABC . If AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ.
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Hint

$\dfrac{AP}{PB}=\dfrac{1}{3}\\ \\ \dfrac{AQ}{AC}=\dfrac{1}{3}$
$\triangle APQ\sim\triangle ABC$ (SS)
$\dfrac{AP}{PB}=\dfrac{AQ}{AC}=\dfrac{PQ}{BC}\\ \\ \dfrac{PQ}{BC}=\dfrac{1}{3}\Rightarrow 3PQ=BC$
Question 10 of 38

10. Question
1 point(s)
In a triangle ABC, altitudes AL and BM intersect in O. Prove that $\dfrac{AO}{BO}=\dfrac{OM}{OL} $.
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Hint

In $\triangle AOM$ and $\triangle BOL$
$\angle AMO=\angle BLO=VOA$
$\triangle AOM\sim\triangle BOL$ (AA)
$\dfrac{AO}{BO}=\dfrac{OM}{OL}$
Question 11 of 38

11. Question
1 point(s)
A wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
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Hint

$AC^2=AB^2-BC^2\\ \\ =24^2-18^2\\ \\ =576-324=252\\ \\ AC=\sqrt{252}=6\sqrt7\ m$
Question 12 of 38

12. Question
1 point(s)
ABCD is a trapezium with AB||DC. If E and F are on non-parallel sides AD and BC respectively such that EF is parallel to AB, then show that $\dfrac{AE}{ED}=\dfrac{BF}{FC} $.
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Hint

$AB||DC$ and $EF||AB$
So $EF||DC$
In $\triangle ADC$
$EG||DC$
$\dfrac{AE}{ED}=\dfrac{AG}{GC}$ …(1)
In $\triangle CAB$
$AB||GF\\ \\ \dfrac{AG}{GC}=\dfrac{BF}{FC}$ …(2)
From (1) and (2)
$\dfrac{AE}{ED}=\dfrac{BF}{FC}$
Question 13 of 38

13. Question
1 point(s)
In the given figure, OC.OD = OA.OB. Show that $\angle $A = $\angle $C and $\angle $B = $\angle $D.
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Hint

$OA\cdot OB=OC\cdot OD$
$\dfrac{OA}{OC}=\dfrac{OD}{OB}$ …(1)
In $\triangle AOD$ and $\triangle COB$
$\dfrac{OA}{OC}=\dfrac{OD}{OB}$ …(1)
$\angle AOD=\angle COB$ (VOA)
$\triangle AOD\sim\triangle COB$ (SAS)
$\angle A=\angle C,\ \angle B=\angle D$
( $\because$ If two triangles are similar their corresponding angles are equal)
Question 14 of 38

14. Question
1 point(s)
An aeroplane leaves an airport and flies due north at a speed of 500 km per hour. At the same time, another aeroplane leaves the same airport and flies due to west at a speed of 1000 km per hour. How far apart will be the two planes after 2 hours?
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Hint

$AB=1000\times\dfrac{3}{2}=1500\ km\\ \\ BC=1200\times\dfrac{3}{2}=1800\ km\\ \\ AC=\sqrt{AB^2+BC^2}=\sqrt{(1500)^2+(1800)^2}\\ \\ =\sqrt{5490000}=2343\ km$ (approx)
Question 15 of 38

15. Question
1 point(s)
In the figure, A, B and C are points on OP,OQ and OR respectively such that AB||PQ and AC||PR. Show that BC||QR.
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Hint

$AB||PQ$
$\dfrac{OA}{AP}=\dfrac{OB}{BQ}$ (using BPT) ….(1)
$AC||PR$
$\dfrac{OC}{CR}=\dfrac{OA}{AP}$ ….(2) (using BPT)
From (1) and (2)
$\dfrac{OB}{BQ}=\dfrac{OC}{CR}$
$BC||QR$ (converse of BPT)
Question 16 of 38

16. Question
1 point(s)
In the figure, two triangles ABC and DBC are on the same base BC in which $\angle A=\angle D=90^o $. If CA and BD meet each other at E show that $\mathrm{AE\times CE=BE\times DE} $.
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Hint

In $\triangle ABC$ and $\triangle DBC$
$\angle A=\angle D=90^o$
$\angle AEB=\angle DEC$ (VOA)
$\triangle ABC\sim\triangle DBC$ (AA)
$\dfrac{AE}{DE}=\dfrac{BE}{CE}\\ \\ AE\times CE=BE\times DE$
Question 17 of 38

17. Question
1 point(s)
In the figure, PQR and SQR are two triangles on the same base QR. If PS intersect QR at O, then show that : $\dfrac{ar\ (PQR)}{ar\ (SQR)}=\dfrac{PO}{SO} $.
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Hint

$\dfrac{ar(\triangle PQR)}{ar(\triangle SQR)}=\dfrac{\frac{1}{2}\times QR\times AP}{\frac{1}{2}\times QR\times BS}=\dfrac{AP}{BS}$ ….(1)
In $\triangle PAO$ and $\triangle SBO$
$\angle PAO=\angle SBO=90^o$
$\angle POA\sim\triangle SBO$ (AA)
$\dfrac{AP}{BS}=\dfrac{PO}{SO}$ ….(2)
From (1) and (2)
$\dfrac{ar(\triangle PQR)}{ar(\triangle SQR)}=\dfrac{PO}{SO}$
Question 18 of 38

18. Question
1 point(s)
In the figure, $O $ is a point inside $\triangle PQR $ such that $\angle POR=90^o\ OP=6\ cm $ and $OR=8\ cm $. If $PR=24\ cm,QR=26\ cm $, prove that $\triangle QPR $ is a right angled triangle.
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Hint

In $\triangle POR$
$PR^2=PO^2+OR^2=6^2+8^2=100\\ \\ PR=10\ cm\\ \\ PQ^2=24^2=576\\ \\ QR^2=26^2=676\\ \\ PQ^2+PR^2=676=QR^2$
$\angle QPR=90^o$ (by converse of pythagoras theorem)
Hence $\triangle QPR$ is a height angled triangle.
Question 19 of 38

19. Question
1 point(s)
Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.
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Hint

$\triangle BCE\sim\triangle ACF$ (AAA similar condition)
$\dfrac{ar\ (\triangle BCE)}{ar\ (\triangle ACF)}=\dfrac{BC^2}{AC^2}\\ \\ =\dfrac{BC^2}{(\sqrt2\ BC)^2}=\dfrac{1}{2}\\ \\ ar\ (\triangle BCE)=\dfrac{1}{2}\ ar\ (\triangle ACF)$
Question 20 of 38

20. Question
1 point(s)
In the figure, DE || AC and DF || AE. Prove that $\dfrac{EF}{BF}=\dfrac{EC}{BE} $.
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Hint

In $\triangle ABC$
$DE||AC\\ \\ \dfrac{BE}{EC}=\dfrac{BD}{DA}$ ….(1)
In $\triangle AEB$
$DF||AE\\ \\ \dfrac{BF}{FE}=\dfrac{BD}{DA}$ ….(2)
From (1) and (2)
$\dfrac{BF}{FE}=\dfrac{BE}{EC}=\dfrac{EF}{BF}=\dfrac{EC}{BE}$ (Proved)
Question 21 of 38

21. Question
1 point(s)
In the figure, if PQ || CB and PR || CD, prove that $\dfrac{AR}{AD}=\dfrac{AQ}{AB} $
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Hint

In $\triangle ABC$
$PQ||BC\\ \\ \dfrac{AQ}{AB}=\dfrac{AP}{AC}$ ….(1) (by basic proportionality theorem)
In $\triangle ACD$
$PR||CD\\ \\ \dfrac{AP}{AC}=\dfrac{AR}{AD}$ …(2)
From (1) and (2)
$\dfrac{AQ}{AB}=\dfrac{AR}{AD}$
Question 22 of 38

22. Question
1 point(s)
In the figure, PM = 6 cm, MR = 8 cm and QR = 26 cm, find the length of PQ.
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Hint

$PR=\sqrt{(PM)^2+(MR)^2}\\ \\ =\sqrt{6^2+8^2}=\sqrt{100}=10\ cm\\ \\ PQ=\sqrt{QR^2-PR^2}=\sqrt{26^2-10^2}=\sqrt{576}=24\ cm$
Question 23 of 38

23. Question
1 point(s)
In the figure, $ABCD $ is a rhombus. Prove that $4AB^2=AC^2+BD^2 $.
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Hint

In $\triangle AOB\ \angle O=90^o$
$AB^2=OA^2+OB^2\\ \\ AB^2=\dfrac{AC^2}{4}+\dfrac{BD^2}{4}\\ \\ 4AB^2=AC^2+BD^2$
Question 24 of 38

24. Question
1 point(s)
In the figure, D,E,F, are mid-point of sides BC, CA, AB respectively of $\triangle $ABC. Find the ratio of area of $\triangle $DEF to area of $\triangle $ABC.
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Hint

We know that the line segment joining the mid points of any two sides of a triangle is half the third side and parallel to it.
$FE=\dfrac{1}{2}BC,\ DE=\dfrac{1}{2}AB,\ FD=\dfrac{1}{2}AC$
In $\triangle ABC$ and $\triangle EFD$
$\dfrac{AB}{ED},\dfrac{BC}{EF},\dfrac{AC}{FD}=2$ ….(1)
$\triangle ABC\sim\triangle EFD$ (SSS)
$\dfrac{ar\ (\triangle ABC)}{ar\ (\triangle EFD)}=\dfrac{(AB)^2}{(ED)^}=4\\ \\ \dfrac{ar\ (\triangle EFD)}{ar\ (\triangle ABC)}=\dfrac{1}{4}$
Question 25 of 38

25. Question
1 point(s)
If one diagonal of a trapezium divides the other diagonal in the ratio 1 : 2, prove that one of the parallel sides is double the other.
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Hint

In $\triangle AOB$ and $\triangle COD$
$\angle AOB=\angle COD$ (VOA)
$\angle ABO=\angle OCD$ (alt $\angle S$ )
$\triangle AOB\sim\triangle DOC$ (AA)
$\dfrac{AO}{OD}=\dfrac{AB}{CD}\\ \\ \Rightarrow \dfrac{1}{2}=\dfrac{AB}{CD}\\ \\ \Rightarrow CD=2AB$
Question 26 of 38

26. Question
1 point(s)
In the figure DE || BC. Find x.
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Hint

$\dfrac{4x-3}{3x-1}=\dfrac{8x-7}{5x-3}\\ \\ 4x(5x-3)-3(5x-3)=8x(3x-1)-7(3x-1)\\ \\ \Rightarrow 20x^2-12x-15x+9=24x^2-8x-21x+7\\ \\ \Rightarrow -4x^2-27x+2=-29x\\ \\ \Rightarrow -4x^2+2x+2=0\\ \\ \Rightarrow 2x^2-x-1=0\\ \\ \Rightarrow 2x^2-2x+x-1=0\Rightarrow 2x(x-1)+1(x-1)=0\\ \\ \Rightarrow x=1,\dfrac{-1}{2}\Rightarrow x=1$
Question 27 of 38

27. Question
1 point(s)
In the figure, $AB||DE $ and $BD||EF $. Prove that $DC^2=CF\times AC $.
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Hint

In $\triangle CDB,\ EF||BD$
$\dfrac{CF}{CD}=\dfrac{CE}{BC}$ (By BPT) …(1)
In $\triangle CAB,\ DE||AB$
$\dfrac{DC}{AC}=\dfrac{CE}{BC}$ (By BPT) …(2)
From (1) and (2)
$\dfrac{CF}{CD}=\dfrac{DC}{AC}\Rightarrow DC^2=CF\times AC$
Question 28 of 38

28. Question
1 point(s)
If $D,E $ are points on the sides $AB $ and $AC $ of a $\triangle ABC $ such that $AD=6\ cm,\ BD=9\ cm,\ AE=8\ cm\ EC=12\ cm $. Prove that $DE||BC $.
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Hint

$\dfrac{AD}{BD}=\dfrac{AE}{EC}\\ \\ \dfrac{6}{9}=\dfrac{8}{12}\\ \\ \dfrac{2}{3}=\dfrac{2}{3}$
By converse of BPT
$DE||BC$
Question 29 of 38

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

In $\triangle PQO$
$DE||OQ\\ \\ \dfrac{PE}{EQ}=\dfrac{PD}{DO}$ ….(1)
In $\triangle PRO$
$DF||OR\\ \\ \dfrac{PF}{FR}=\dfrac{PD}{DO}$ ….(2)
From (1) and (2)
$\dfrac{PE}{EQ}=\dfrac{PF}{FR}\quad EF||QR$
Question 30 of 38

30. Question
1 point(s)
In the figure $\angle ACB=90^o $ and $CD\bot AB $. Prove that $\dfrac{BC^2}{AC^2}=\dfrac{BD}{AD} $.
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Hint

If a perpendicular is drawn from the vertex of the right angle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
$\triangle ACD\sim\triangle ABC\\ \\ \triangle BCD\sim\triangle BAC\\ \\ \dfrac{AC}{AB}=\dfrac{AD}{AC}\Rightarrow AC^2=AB\times AD$ ..(1)
Also $\dfrac{BC}{BA}=\dfrac{BD}{BC}$
$\Rightarrow BC^2=BA\times BD$ ….(2)
Dividing eqn (2) by (1)
$\dfrac{BC^2}{AC^2}=\dfrac{BA\times BD}{AB\times AD}\\ \\ \Rightarrow \dfrac{BC^2}{AC^2}=\dfrac{BD}{AD}$ (Proved)
Question 31 of 38

31. Question
1 point(s)
In the figure, $XN || CA$ and $XM || BA.\ T$ is a point on $CB$ produced. Prove that $TX^2=TB.TC. $
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Hint

In $\triangle TXM,\ BN||XM$
$\dfrac{TB}{TX}=\dfrac{TN}{TM}$ (By BPT) ….(1)
In $\triangle TCM,\ XN||CM$
$\dfrac{TX}{TC}=\dfrac{TN}{TM}$ (by BPT) ….(2)
From (1) and (2)
$\dfrac{TB}{TX}=\dfrac{TX}{TC}\Rightarrow TX^2=TB\cdot TC$
Question 32 of 38

32. Question
1 point(s)
In the figure. $AB\bot BC,\ DE\ \bot\ AC $ and $GF\ \bot\ BC $. Prove that $\triangle ADE\sim\triangle GCF $.
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Hint

In $\triangle ABC$ and $\triangle ADE$
$\angle A=\angle A\\ \\ \angle DAE=\angle BAC$
$\triangle ABC\sim\triangle ADE$ (AA)
In $\triangle ABC$ and $\triangle CFD$
$\angle C=\angle C\\ \\ \angle GFC=\angle ABC$
$\triangle ABC\sim\triangle GFC$ (AA)
So $\triangle ADE\sim\triangle GFC$
Question 33 of 38

33. Question
1 point(s)
From the given figure, find $\angle $MLN.
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Hint

$\dfrac{AB}{LM}=\dfrac{BC}{MN}=\dfrac{AC}{LN}=\dfrac{2}{5}$
$\triangle ABC\sim\triangle LMN$ (SSS)
$\angle MLN=\angle BAC=180-(50+70)\\ \\ =180-120=60^o$
Question 34 of 38

34. Question
1 point(s)
In a quadrilateral ABCD, $\angle B=90^o $. If $AD^2=AB^2+BC^2+CD^2 $, prove that $ \angle ACD=90^o$.
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Hint

$AD^2=AB^2+BC^2+CD^2$ (given)
But $AB^2+BC^2=AC^2$ (by pythagoras theorem)
$AD^2=AC^2+CD^2$
By converse of pythagoras theorem
$\angle ACD=90^o$
Question 35 of 38

35. Question
1 point(s)
$ABC $ is an isosceles triangle with $AC=BC$. If $AB^2=2AC^2 $, prove that $ABC $ is a right angled triangle.
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Hint

$AB^2=2AC^2\\ \\ =AC^2+AC^2\\ \\ AB^2=AC^2+BC^2\ (\because AC=BC)$
So $AB$ is the hypotenuse
$(Hypotenuse)^2=(Height)^2+(Base)^2\\ \\ =AC^2+BC^2=2\ AC^2\\ \\ AB^2=AB^2$
Hence $\triangle ABC$ is a right angled triangle
Question 36 of 38

36. Question
1 point(s)
D,E and F are respectively the mid-point of sides AB, BC and CA of $\triangle $ABC . Determine the ratio of the areas of $\triangle $DEF and $\triangle $ABC .
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Hint

$FE||AB$
$FE=\dfrac{1}{2}AB$ (by mid point theorem)
$FE||BD$ and $FE=BD$
So $FEBD$ is a parallelogram
Similarly $DFCE$ is a parallelogram
$\dfrac{ar\ (\triangle DEF)}{ar\ (\triangle ABC)}=\dfrac{DE^2}{AD^2}=\dfrac{\left(\dfrac{1}{2}AB\right)^2}{AD^2}=\dfrac{1}{4}$
Question 37 of 38

37. Question
1 point(s)
In the given figure, $PQ=24\ cm,\ QR=26\ cm,\ \angle PAR=90^o,\ PA=6\ cm $ and $AR=8\ cm $. Find $\angle QPR $.
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Hint

$PR^2=AR^2+AP^2=8^2+6^2=100\\ \\ PR=10\ cm$
In $\triangle PQR$
$PQ^2+PR^2=(24)^2+(10)^2=576+600=676=QR^2$
Hence by converse of pythagoras theorem $\angle QPR=90^o$
Question 38 of 38

38. Question
1 point(s)
A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point?
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Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\sqrt{15^2+8^2}=\sqrt{225+64}\\ \\ =\sqrt{289}=17\ m$
AC is 17m from his starting point