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Geometry and Mensuration: Test 14

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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:
geometry-and-mensuration-test-14-qestion-1-pic-1
Question 2
The length of the diagonal BD of the parallelogram ABCD is 18 cm. If P and Q are the centroid of the ΔABC and ΔADC 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:
10
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 Δ( BCP)= areaΔ ( DPQ)
B
area Δ( BCP) >areaΔ ( DPQ)
C
area Δ( BCP)
D
area Δ( BCP) +areaΔ( DPQ)
Question 3 Explanation:
geometry-and-mensuration-test-14-qestion-2-pic-1
We know that: Triangles ΔAPC and ΔBCP have the common base PC.
Also, the two triangles ΔAPC and ΔBCP are located between the same two parallel lines: AB and PC
Thus, we can say: AD||CQ and AD=CQ……………………….i
From the above, we can conclude ADQC is a parallelogram.
Also, triangles ΔADC and ΔDAQ have the common base: AD
Also, the two triangles ΔADC and ΔDAQ are located between the same two parallel lines: AD and CQ.
Thus, we can conclude: Area of Δ( ADC)= Area of Δ( ADQ)
Let’s subtract area of ( DADP) from both sides: area of Δ( APC)= area of Δ( DPQ)……..(ii)
From (i) and (ii), we can conclude: area of Δ( BPC) = area of Δ( DPQ)
Question 4
ABCD is a rhombus. A straight line through C cuts AD produced at P and AB produced at Q. if DP= ½ 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:
11
geometry-and-mensuration-test-14-qestion-4-pic-1
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:
12
$ \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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