Mathematics Class IX

Question 1 of 15

1. Question
1 point(s)
The perimeter of a triangle is 36 cm. One side of the triangle is 15 cm. Difference of the other two sides is 1 cm. What is the area of this triangle?
- $ 6\sqrt{105} \, cm^{2} $
- $ 12\sqrt{35} \, cm^{2} $
- $ 14\sqrt{21} \, cm^{2} $
- $ 15\sqrt{14} \, cm^{2} $
Correct

Incorrect

Hint

Let $a=11$ and the other two sides be $b$ and $c$ .

(1) $\begin{equation*}\begin{split} 11+b+c &= 36 \\ b+c &= 25 \\ b-c &= 1 \end{split}\end{equation*}$

Solving the above two equations we get $b=13, c=12$ .

(2) $\begin{equation*}\begin{split} s &= \dfrac{a+b+c}{2} \\ s &= \dfrac{11+13+12}{2} \\ s &= 18 \\ Area \, of \, triangle &= \sqrt{s(s-a)(s-b)(s-c)} \\ &= \sqrt{18(18-11)(18-13)(18-12)} \\ &= \sqrt{18(7)(5)(6)} \\ &= 6\sqrt{105} \end{split}\end{equation*}$
Question 2 of 15

2. Question
1 point(s)
All sides of a scalene triangle are doubled. What is the percentage increase in the area of the triangle?
- 4
- 3
- 2
- 1
Correct

Incorrect

Hint

Let the sides of a triangle be $a, b, c$ and the sides of the new triangle will be $2a, 2b, 2c$ . Semi-perimeter of old triangle is $S=\dfrac{a+b+c}{2}$ . Semi-perimeter of new triangle is $\dfrac{2a+2b+2c}{2}$ which is two times the semi-perimeter of old triangle.

(1) $\begin{equation*}\begin{split} Area \,of \,old \,triangle &= \sqrt{s(s-a)(s-b)(s-c)} \\ Area\,of\,new\,triangle &= \sqrt{2s(2s-2a)(2s-2b)(2s-2c)} \\ &= 4 \times area\,of\,old\,triangle \end{split}\end{equation*}$

Therefore, increase in the area is 300%.
Question 3 of 15

3. Question
1 point(s)
A triangular parking space with dimensions 40 m X 50 m X 60 m is to be painted. The cost of painting is Rs 15 per square meter. Find the total cost of painting.
- $ Rs \, 375\sqrt{5} $
- $ Rs \, 3750\sqrt{7} $
- $ Rs \, 5625\sqrt{7} $
- $ Rs \, 4250\sqrt{5} $
Correct

Incorrect

Hint

(1) $\begin{equation*}\begin{split} s &= \dfrac{a+b+c}{2} \\ s &= \dfrac{40+50+60}{2} \\ s &= 75 \\ Area \, of \, triangle &= \sqrt{s(s-a)(s-b)(s-c)} \\ &= \sqrt{75(75-40)(75-50)(75-60)} \\ &= 375\sqrt{7} \\ Cost\,of\,painting &= 15\times375\sqrt{7} \\ &= Rs \, 5625\sqrt{7} \end{split}\end{equation*}$
Question 4 of 15

4. Question
1 point(s)
What is the area of the quadrilateral ABCD if AB = 10 cm, BC = 10 cm, CD = 12 cm, AD = 12 cm and the diagonal AC = 12 cm?
- $ 12(4\sqrt{6}+3) \, cm^{2} $
- $ 12(3\sqrt{6}+4) \, cm^{2} $
- $ 12(4\sqrt{3}+3) \, cm^{2} $
- $ 12(3\sqrt{3}+4) \, cm^{2} $
Correct

Incorrect

Hint

(1) $\begin{equation*}\begin{split} Area\,of\,equilateral\,triangle\,ADC &= \dfrac{\sqrt{3}}{4}12^{4} \\ &= 36\sqrt{3} cm^{2} \\ s\, for\,triangle\,ABC &= \dfrac{10+10+12}{2} \\ &= 16 \\ Area \, of \, triangle\, ABC &= \sqrt{16(16-10)(16-10)(16-12)} \\ &= 48 \, cm^{2} \\ Area\,ABCD&= 36\sqrt{3}+48 \\ &= 12(3\sqrt{3}+4) \, cm^{2} \end{split}\end{equation*}$
Question 5 of 15

5. Question
1 point(s)
What is the area of the quadrilateral ABCD if AB = 16 cm, BC = 30 cm, CD = 18 cm, AD = 18 cm, the diagonal AC = 34 cm and the angle ABC is 90°?
- $ 240+16\sqrt{35} \, cm^{2} $
- $ 240+17\sqrt{35} \, cm^{2} $
- $ 360+35\sqrt{17} \, cm^{2} $
- $ 300+17\sqrt{35} \, cm^{2} $
Correct

Incorrect

Hint

(1) $\begin{equation*}\begin{split} Area\,of\,triangle\,ABC &= \dfrac{1}{2}(16)(30) \\ &= 240\, cm^{2} \\ s\, for\,triangle\,ADC &= \dfrac{18+18+34}{2} \\ &= 35 \\ Area \, of \, triangle\, ADC &= \sqrt{35(35-18)(35-18)(35-34)} \\ &= 17\sqrt{35} \, cm^{2} \\ Area\,ABCD&= 240+17\sqrt{35} \, cm^{2} \end{split}\end{equation*}$
Question 6 of 15

6. Question
1 point(s)
The perimeter of a rhombus is 160 cm. If one of the diagonals is 60 cm long, then what is the area of rhombus?
- $ 300\sqrt{6} \, cm^{2} $
- $ 600\sqrt{6} \, cm^{2} $
- $ 300\sqrt{3} \, cm^{2} $
- $ 600\sqrt{3} \, cm^{2} $
Correct

Incorrect

Hint

Perimeter of rhombus is 160 cm so it means its side length is 40 cm. The diagonal divides the rhombus in two triangles of equal areas.

(1) $\begin{equation*}\begin{split} s\, for\,triangle &= \dfrac{40+40+60}{2} \\ &= 70 \\ Area \, of \, triangle &= \sqrt{70(70-40)(70-40)(70-60)} \\ &= 300\sqrt{7} \, cm^{2} \\ Area\, of\, rhombus &= 600\sqrt{3} \, cm^{2} \end{split}\end{equation*}$
Question 7 of 15

7. Question
1 point(s)
A floral design is made with 8 isosceles triangles with base 10 cm and equal sides as 15 cm each. Find the area of the design.
- $ 400\sqrt{2} \, cm^{2} $
- $ 400\sqrt{3} \, cm^{2} $
- $ 600\sqrt{2} \, cm^{2} $
- $ 200\sqrt{6} \, cm^{2} $
Correct

Incorrect

Hint

(1) $\begin{equation*}\begin{split} s\, for\,triangle &= \dfrac{10+15+15}{2} \\ &= 20 \\ Area \, of \, triangle &= \sqrt{20(20-10)(20-15)(20-15)} \\ &= 50\sqrt{2} \, cm^{2} \\ Area\, of\,design &= 8\times 50\sqrt{2} \\ &= 400\sqrt{2} \, cm^{2} \end{split}\end{equation*}$
Question 8 of 15

8. Question
1 point(s)
Area of an equilateral triangle is $ 64\sqrt{2} \, cm^{2} $. What is the length of each side of the triangle?
- 20 cm
- 12 cm
- 8 cm
- 16 cm
Correct

Incorrect

Hint

Let the length of the side of the equilateral triangle be $a$ .

(1) $\begin{equation*}\begin{split} Area\,of\,an\,equilateral\,triangle &= \dfrac{\sqrt{3}}{4}a^{2} \\ 64\sqrt{3} &= \dfrac{\sqrt{3}}{4}a^{2} \\ 16 &= a \end{split}\end{equation*}$
Question 9 of 15

9. Question
1 point(s)
The perimeters of a square and an equilateral triangle are same. The diagonal of the square is $ 18\sqrt{2} \, cm$. What is the area of the equilateral triangle?
- $ 192\sqrt{2} \, cm^{2} $
- $ 288\sqrt{2} \, cm^{2} $
- $ 144\sqrt{3} \, cm^{2} $
- $ 96\sqrt{2} \, cm^{2} $
Correct

Incorrect

Hint

The diagonal of the square is $18\sqrt{2} \, cm$ . Side of the square is 18 cm. Perimeter of the triangle is 72 cm. Side of the triangle is 24 cm.

(1) $\begin{equation*}\begin{split} Area\,of\,an\,equilateral\,triangle &= \dfrac{\sqrt{3}}{4}a^{2} \\ &= \dfrac{\sqrt{3}}{4}24^{2} \\ &= 144\sqrt{3} \end{split}\end{equation*}$
Question 10 of 15

10. Question
1 point(s)
The adjacent sides of a parallelogram 15 cm and 20 cm. If one of the diagonals is 25 cm long, then what is the area of the parallelogram?
- $ 360 \, cm^{2} $
- $ 240 \, cm^{2} $
- $ 300 \, cm^{2} $
- $ 150 \, cm^{2} $
Correct

Incorrect

Hint

The diagonal divides the parallelogram in two triangles of equal areas.

(1) $\begin{equation*}\begin{split} s\, for\,triangle &= \dfrac{15+20+25}{2} \\ &= 30 \\ Area \, of \, triangle &= \sqrt{30(30-15)(30-20)(30-25)} \\ &= 150 \, cm^{2} \\ Area\, of\, parallelogram &= 300 \, cm^{2} \end{split}\end{equation*}$
Question 11 of 15

11. Question
1 point(s)
All sides of a scalene triangle are halved. What is the percentage decrease in the area of the triangle?
- 0.75
- 0.5
- 0.25
- 0.33
Correct

Incorrect

Hint

Let the sides of a triangle be $a, b, c$ and the sides of the new triangle will be $\dfrac{a}{2}, \dfrac{b}{2}, \dfrac{c}{2}$ . Semi-perimeter of old triangle is $S=\dfrac{a+b+c}{2}$ . Semi-perimeter of new triangle is $\dfrac{\frac{a}{2}+\frac{b}{2}+\frac{c}{2}}{2}$ which is half of the semi-perimeter of old triangle.

(1) $\begin{equation*}\begin{split} Area \,of \,old \,triangle &= \sqrt{s(s-a)(s-b)(s-c)} \\ Area\,of\,new\,triangle &= \sqrt{\dfrac{s}{2}(\dfrac{s}{2}-\dfrac{a}{2})(\dfrac{s}{2}-\dfrac{b}{2})(\dfrac{s}{2}-\dfrac{c}{2})} \\ &= \dfrac{1}{4} \times area\,of\,old\,triangle \end{split}\end{equation*}$

So if we assume the old area as 100 then new area will be 25. Therefore, decrease in the area is 75%.
Question 12 of 15

12. Question
1 point(s)
ABCD is a trapezium in which AB = 100 cm, BC = 50 cm, CD = 30 cm and AD = 40 cm. AB and CD are the parallel sides. What is the area of this trapezium?
- $ \dfrac{2400\sqrt{6}}{7} \, cm^{2} $
- $ \dfrac{5200\sqrt{6}}{7} \, cm^{2} $
- $ \dfrac{4800\sqrt{6}}{7} \, cm^{2} $
- $ \dfrac{2600\sqrt{6}}{7} \, cm^{2} $
Correct

Incorrect

Hint

From the point C, draw a line parallel to AD intersecting AB at E. BE will be 70 cm and EC will be 40 cm. Draw a perpendicular from point C on AB which intersects AB at F.

(1) $\begin{equation*}\begin{split} s\, for\,triangle\,BEC&= \dfrac{40+50+70}{2} \\ &= 80 \\ Area \, of \, triangle\,BEC &= \sqrt{80(80-40)(80-50)(80-70)} \\ &= 400\sqrt{6} \, cm^{2} \\ \dfrac{1}{2} \times BE \times CF &= 400\sqrt{6} \\ CF &= \dfrac{80\sqrt{6}}{7} \\ Area\,of\,trapezium\,ABCD &= \dfrac{1}{2} \times (100+30) \times \dfrac{80\sqrt{6}}{7} \\ &= \dfrac{5200\sqrt{6}}{7} \end{split}\end{equation*}$
Question 13 of 15

13. Question
1 point(s)
A kite is made with a square and an isosceles triangle. Length of diagonal of square is 50 cm. The base of isosceles triangle is 20 cm and the other sides are 15 cm each. What is the ratio of the area of square to the area of the isosceles triangle?
- $ 4\sqrt{6}:1 $
- $ 6\sqrt{5}:1 $
- $ 4\sqrt{5}:1 $
- $ 5\sqrt{5}:1 $
Correct

Incorrect

Hint

Use the formula $\dfrac{1}{2}d^{2}$ to find the area of the square.

(1) $\begin{equation*}\begin{split} Area\, of\,square &= \dfrac{1}{2}50^{2} \\ &= 1250 \, cm^{2} \\ s\, for\,triangle &= \dfrac{15+15+20}{2} \\ &= 25 \\ Area \, of \, triangle &= \sqrt{25(25-15)(25-15)(25-20)} \\ &= 25\sqrt{5} \, cm^{2} \\ \dfrac{Area\,of\,square}{Area\,of\,triangle} &= \dfrac{1250}{50\sqrt{5}} \\ &= \dfrac{5\sqrt{5}}{1} \, OR \, 5\sqrt{5}:1 \end{split}\end{equation*}$
Question 14 of 15

14. Question
1 point(s)
Which of the following is required to use Heron’s formula to find the area of a triangle?
- All three sides of the triangle
- All three angles of the triangle
- All three medians of the triangle
- None of these
Correct

Incorrect

Hint

All three sides of a triangle are required to find the area of the triangle using Heron’s formula.
Question 15 of 15

15. Question
1 point(s)
The area of a right isosceles triangle is 32 square units. What is the perimeter of this triangle?
- $ 16(1+\sqrt{2}) $
- $ 8(2+\sqrt{2}) $
- $ 8(1+\sqrt{2}) $
- $ 16(2+\sqrt{2}) $
Correct

Incorrect

Hint

The sides of a right isosceles triangle are $x, x$ and $\sqrt{2} x$ .

(1) $\begin{equation*}\begin{split} \dfrac{1}{2}x^{2} &= 32 \\ x &= 8 \\ Perimeter &= 8+8+8\sqrt{2} \\ &= 8(2+\sqrt{2}) \end{split}\end{equation*}$