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• This is an assessment test.
• These tests focus on geometry and mensuration and are meant to indicate your preparation level for the subject.
• Kindly take the tests in this series with a pre-defined schedule.

## Geometry and Mensuration: Test 28

Congratulations - you have completed Geometry and Mensuration: Test 28. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%%
 Question 1
ABCD is a rhombus. AB is produced to F and BA is produced to E such that AB= AE= BF. Also ED and CF are produced to meet at G. Then:
 A ED > CF B EGâŠ¥GF C ED2 +CF2= EF2 D ED||CF
Question 1 Explanation:
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Here AB= AE= BF.......(1) Given ABCD is a rhombus.
Therefore, AB = BC = CD = AD ... (2)
On equating (1) and (2), we get
BC = BF
â‡’âˆ 4 = âˆ 3 [Angles opposite to equal sides are equal]
Again, âˆ B is the exterior angle of triangle BFC.
Therefore, âˆ 2 = âˆ 3 + âˆ 4 = 2âˆ 4 ... (3)
Similarly, AE = AD â‡’âˆ 5 = âˆ 6
Also, âˆ A is the exterior angle of triangle ADE
â‡’âˆ 1 = 2âˆ 6 ... (4)
Also, âˆ 1 +âˆ 2 = 180Â° [consecutive interior angles]
âˆ´ 2âˆ 6 + 2âˆ 4 = 180Â°
â‡’âˆ 6 + âˆ 4 = 90Â° ... (5)
Now, in triangle EGF, by angle sum property of triangle
â‡’ 90Â°+ âˆ G = 180Â° [using (5)]
Thus, EG âŠ¥ FG.
 Question 2
Two equal circles of radius 4 cm intersect each other such that each passes through the centre of the other. The length of the common chord is:
 A $\displaystyle 2\sqrt{3}$ B $\displaystyle 4\sqrt{3}$ C $\displaystyle 2\sqrt{2}$ D $\displaystyle 8$
Question 2 Explanation:
$\begin{array}{l}The\text{ }length\text{ }of\text{ }the\text{ }common\text{ }chord\text{ }=\text{ }\\2\text{ }\times \text{ }length\text{ }of\text{ }the\text{ }altitude\text{ }of\text{ }the\text{ }equilateral\text{ }triangle\text{ }of\text{ }side\text{ }4\text{ }cm.\\Thus\text{ }the\text{ }length\text{ }of\text{ }common\text{ }chord\text{ }is\\2\times \frac{\sqrt{3}}{2}\times 4\\=4\sqrt{3}\end{array}$
 Question 3
The length of two chords AB and AC of a circle are 8 cm and 6 cm and âˆ BAC= 90o, then the radius of circle is
 A 25 cm B 20 cm C 4 cm D 5 cm
Question 3 Explanation:

$\begin{array}{l}The\text{ }diameter=\sqrt{{{8}^{2}}+{{6}^{2}}}=10\\Thus\text{ }the\text{ }radius\text{ }=5.\end{array}$
 Question 4
The tangents are drawn at the extremities of a diameter AB of a circle with centre P. If a tangent to the circle at the point C intersects the other two tangents at Q and R, then the measure of the âˆ QPR is
 A 45o B 60o C 90o D 180o
Question 4 Explanation:

âˆ QCP= 900,
âˆ QAP=900.
AQ=QC.
Thus AQCP is a square.
âˆ QPC= 450
Similarly âˆ RPC=450
Thus âˆ QPR = 900.
Correct option is (c)
 Question 5
The circumcentre of a triangle Î”ABC is O. If âˆ BAC= 85o and BCA= 75o, then the value of âˆ OAC is
 A 40o B 60o C 70o D 90o
Question 5 Explanation:

âˆ BAC=85, âˆ BCA=75
âˆ AOC=360-2(BAC+BCA)
=360 â€“ 320=40
Thus âˆ OAC= Â½ (180-40)=70
Correct option is (c)
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