RD Chapter 14 Quadrilaterals Ex VSAQS Solutions
Question - 1 : - In a parallelogram ABCD, write the sum of angles A and B.
Answer - 1 : -
In parallelogram ABCD, Adjacent angles of a parallelogram are supplementary.
Therefore, ∠A + ∠B = 1800
Question - 2 : - In a parallelogram ABCD, if ∠D = 1150, then write the measure of ∠A.
Answer - 2 : -
In a parallelogramABCD,
∠D = 1150 (Given)
Since, ∠A and ∠D are adjacent angles of parallelogram.
We know, Adjacentangles of a parallelogram are supplementary.
∠A + ∠D = 1800
∠A = 1800 –1150 = 650
Measure of ∠A is 650.
Question - 3 : - PQRS is a square such that PR and SQ intersect at O. State the measure of ∠POQ.
Answer - 3 : -
PQRS is a square suchthat PR and SQ intersect at O. (Given)
We know, diagonals ofa square bisects each other at 90 degrees.
So, ∠POQ = 900
Question - 4 : - In a quadrilateral ABCD, bisectors of angles A and B intersect at O such that ∠AOB = 75°, then write the value of ∠C + ∠D.
Answer - 4 : -
∠AOB = 75o (given)
In a quadrilateralABCD, bisectors of angles A and B intersect at O, then
∠AOB = 1/2 (∠ADC + ∠ABC)
or ∠AOB = 1/2 (∠D + ∠C)
By substituting givenvalues, we get
75 o =1/2 (∠D + ∠C)
or ∠C + ∠D = 150 o
Question - 5 : - The diagonals of a rectangle ABCD meet at O. If ∠BOC = 44o, find ∠OAD.
Answer - 5 : -
ABCD is a rectangleand ∠BOC = 44o (given)
∠AOD = ∠BOC (vertically opposite angles)
∠AOD = ∠BOC = 44o
∠OAD = ∠ODA (Angles facing same side)
and OD = OA
Since sum of all theangles of a triangle is 180 o, then
So, ∠OAD = 1/2 (180 o – 44 o)= 68 o
Question - 6 : - If PQRS is a square, then write the measure of ∠SRP.
Answer - 6 : -
PQRS is a square.
⇒ All side are equal,and each angle is 90o degrees and diagonals bisect the angles.
So, ∠SRP = 1/2 (90 o) = 45o
Question - 7 : - If ABCD is a rectangle with ∠BAC = 32o,find the measure of ∠DBC.
Answer - 7 : -
ABCD is a rectangleand ∠BAC=32 o (given)
We know, diagonals ofa rectangle bisects each other.
AO = BO
∠DBA = ∠BAC = 32 o (Angles facing same side)
Each angle of arectangle = 90 degrees
So, ∠DBC + ∠DBA = 90 o
or ∠DBC + 32 o = 90 o
or ∠DBC = 58 o
Question - 8 : - If ABCD is a rhombus with ∠ABC = 56o,find the measure of ∠ACD.
Answer - 8 : -
In a rhombus ABCD,
o
So,
or
We know, consecutiveangles of a rhombus are supplementary.
∠BCD + ∠ABC = 180 o
∠BCD = 180 o – 56 o = 124 o
Equation (1) ⇒ o = 62 o
Question - 9 : - The perimeter of a parallelogram is 22 cm. If the longer side measure 6.5cm, what is the measure of shorter side?
Answer - 9 : -
Perimeter of aparallelogram = 22 cm. (Given)
Longer side = 6.5 cm
Let x be the shorterside.
Perimeter = 2x + 2×6.5
22 = 2x + 13
2x = 22 – 13 = 9
or x = 4.5
Measure of shorterside is 4.5 cm.
Question - 10 : - If the angles of a quadrilateral are in the ratio 3:5:9:13, then find the measure of the smallest angle.
Answer - 10 : -
Angles of aquadrilateral are in the ratio 3 : 5 : 9 : 13 (Given)
Let the sides are 3x,5x, 9x, 13x
We know, sum of allthe angles of a quadrilateral = 360o
3x + 5x + 9x + 13x =360 o
30 x = 360 o
x = 12 o
Measure of smallestangle = 3x = 3(12) = 36o .