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## Geometry and Mensuration: Test 14

Congratulations - you have completed Geometry and Mensuration: Test 14.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
 Question 1
The difference between the exterior and interior angles at a vertex of a regular polygon is 150o. The number of sides of the polygon is
 A 10 B 15 C 24 D 30
Question 1 Explanation:
 Question 2
The length of the diagonal BD of the parallelogram ABCD is 18 cm. If P and Q are the centroid of the Î”ABC and Î”ADC respectively then the length of the line segment PQ is
 A 4 cm B 6 cm C 9 cm D 12 cm
Question 2 Explanation:

Since the points P and Q are the centroids, therefore PQ = 1/3 of BD.
Thus PQ = 1/3 of 18 cm = 6 cm.
 Question 3
ABCD is a parallelogram where AB is parallel to CD. BC is produced to Q such that BC= CQ, where P is the point on DC
 A area Î”( BCP)= areaÎ” ( DPQ) B area Î”( BCP) >areaÎ” ( DPQ) C areaÂ Î”( BCP) D area Î”( BCP) +areaÎ”( DPQ)
Question 3 Explanation:

We know that: Triangles Î”APC and Î”BCP have the common base PC.
Also, the two triangles Î”APC and Î”BCP are located between the same two parallel lines: AB and PC
From the above, we can conclude ADQC is a parallelogram.
Also, the two triangles Î”ADC and Î”DAQ are located between the same two parallel lines: AD and CQ.
Thus, we can conclude: Area of Î”( ADC)= Area of Î”( ADQ)
Letâ€™s subtract area of ( DADP) from both sides: area of Î”( APC)= area of Î”( DPQ)â€¦â€¦..(ii)
From (i) and (ii), we can conclude: area of Î”( BPC) = area of Î”( DPQ)
 Question 4
ABCD is a rhombus. A straight line through C cuts AD produced at P and AB produced at Q. if DP= Â˝ AB, Then the ratio of the lengths of BQ and AB is
 A 2: 1 B 1: 2 C 1: 1 D 3: 1
Question 4 Explanation:

 Question 5
In a quadrilateral ABCD, with unequal sides if the diagonals AC and BD intersect at right angles, then
 A AB2 +BC2= DC2 +DA2 B AB2 +CD2 = BC2+ DA2 C AB2 +AD2= BC2+ CD2 D AB2+ BC2 =2(CD2 +DA2)
Question 5 Explanation:

$\begin{array}{l}Let\text{ }the\text{ }diagonals\text{ }intersect\text{ }at\text{ }O.\\A{{O}^{2}}+B{{O}^{2}}=A{{B}^{2}}~,~~\\B{{O}^{2}}+C{{O}^{2}}=B{{C}^{2}}~\\C{{D}^{2}}=O{{C}^{2}}+O{{D}^{2}}\\A{{D}^{2}}=A{{O}^{2}}+A{{D}^{2}}\\From\text{ }the\text{ }equations\text{ }we\text{ }can\text{ }check\text{ }the\text{ }options\text{ }.\\A{{B}^{2}}+C{{D}^{2}}=B{{C}^{2}}+D{{A}^{2}}=B{{O}^{2}}+O{{D}^{2}}+A{{O}^{2}}+C{{O}^{2}}\\We\text{ }can\text{ }conclude\text{ }that\text{ }correct\text{ }option\text{ }is\text{ }\left( b \right)\end{array}$
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