Geometry and Mensuration: Test 23

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Geometry and Mensuration: Test 23

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Question 1

If the medians of two equilateral triangles are in the ratio 1:2 then what is the ratio of their sides?

A	1: 2
B	2: 3
C	3: 2
D	√3: √2

Question 1 Explanation:

$\displaystyle \begin{array}{l}Median\text{ }of\text{ }an\text{ }equilateral\text{ }triangle=\frac{\sqrt{3}}{2}a\\By\text{ }given\text{ }condition\\\frac{\frac{\sqrt{3}}{2}{{a}_{1}}}{\frac{\sqrt{3}}{2}{{a}_{2}}}=\frac{1}{2}\\\therefore \,\,\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{2}\end{array}$

Question 2

The hypotenuse of a right triangle is 3√10 unit. If the smaller side is tripled and the longer side is doubled, new hypotenuse becomes 9√5 unit. What are the lengths of the smaller and longer sides of the right triangle respectively?

A	5 unit and 9 units
B	5 units and 6 units
C	3 units and 9 units
D	3 units and 6 units

Question 2 Explanation:

Let the sides be x , y where x<y
x²+ y²=(3√10)² = 90
9x²+ 4y²=(9√5)² = 405
Solving the two equations
x = 3 and y=9

Question 3

In the figure given above, PQ = QS and QR = RS. If ∠SRQ = 100°, how many degrees is ∠QPS? cat-geometry-and-mensuration-10-png-1

A	40°
B	30°
C	20°
D	15°

Question 3 Explanation:

Question 4

ABC is a right angled triangle, right angled at C and p is the length of the perpendicular from C on AB. If a, b and c are the lengths of the sides BC, AC and AB respectively, then which one of the following is correct?

A	(a² + b²)p² = a²b²
B	a² + b² = a²b²p²
C	p²= a² + b²
D	p²= a² –b²

Question 4 Explanation:

$\begin{array}{l}\frac{1}{2}pc=\frac{1}{2}ab\,\,\left( Since\text{ }area\text{ }of\text{ }same\text{ }triangles \right)\\{{p}^{2}}{{c}^{2}}={{a}^{2}}{{b}^{2}}\\=>{{p}^{2}}({{a}^{2}}+{{b}^{2}})={{a}^{2}}{{b}^{2}}\end{array}$

Question 5

AB, EF and CD are parallel lines. Given that, EF= 5 cm GC = 10 cm, AB = 15 cm and DC = 18 cm. What is the value of AC?

A	20 cm
B	24 cm
C	25 cm
D	28 cm

Question 5 Explanation:

$\begin{array}{l}\because AB||EF||CD\\\frac{EF}{CD}=\frac{EG}{GC}\\=>\frac{EF}{18}=\frac{5}{10}\\=>EF=9\\In\vartriangle ABC\text{ }and\vartriangle EFC\text{ }\left( similar\text{ }triangles \right),\\\frac{EF}{AB}=\frac{EC}{AC}\\=>9/15=15/AC\\AC=(15\times 15)/9=25cm\end{array}$

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