Class 9th Mathematics Part Ii MHB Solution
Practice Set 5.4
Question 1.In IJKL, side IJ || side KL ∠I = 108° ∠K = 53° then find the measures of ∠J and ∠L.
Answer:
IJ ∥ KL and IL is transversal
∠I + ∠L = 180° (adjacent angles on the same side of the transversal)
⇒ ∠L = 180° – 108° = 72°
Now again IJ ∥ KL and JK is transversal
∠J + ∠K= 180 ° (adjacent angles on the same side of the transversal)
⇒ ∠K = 180° – 53° = 127°
Question 2.
In ABCD, side BC || side AD, side AB ≅ sided DC If ∠A = 72° then find the measures of ∠B, and ∠D.
Answer:
Given that BC ∥ AD and BC = AD (congruent)
⇒ the quadrilateral is a parallelogram (pair of opposite sides are equal and parallel)
∠ A = 72°
⇒ ∠C = 72° (opposite angles of parallelogram are congruent)
∠B = 180° – 72° = 108° (adjacent angles of a parallelogram are supplementary)
∠D = 108° (opposite angles of parallelogram are congruent)
Question 3.
In ABCD, side BC < side AD (Figure 5.32) side BC || side AD and if side BE ≅ side CD then prove that ∠ABC ≅ ∠DCB.
Answer:
The figure of the question is given below:
Construction: we will draw a segment ∥ to BA meeting BC in E through point D.
Given BC ∥ AD
And AB ∥ ED (construction)
⇒ AB = DE (distance between parallel lines is always same)
Hence ABDE is parallelogram
⇒ ∠ABE ≅ ∠DEC (corresponding angles on the same side of transversal)
And segBA ≅ seg DE (opposite sides of a ∥gram)
But given BA ≅ CD
So seg DE ≅ seg CD
⇒∠CED ≅ ∠DCE ( ∵ Δ CED is isosceles with CE = CD)
(Angle opposite to opposite sides are equal)
⇒ ∠ABC ≅ ∠DCB
