(0) DABCD is a trapezium. CD || AB. If DO = 3, CO=x-5, BO=x-3 and AO-3x - 19, then answer the following questions. D 3 x-5 x-3 37-19 A B (a) Prove AAOBACOD. (b) Write the corresponding sides of A AOB and A COD and frame an equation involving x. (c) Find the value of x.
Answers
Answer:
For Fig.a,
AB∥CD [ Given ]
∴ Quadrilateral ABCD is a trapezium.
∴
CO
AO
=
DO
BO
[ Diagonals of trapezium divides each other proportionally ]
⇒
4x−2
4
=
2x+4
x+1
⇒ 4(2x+4)=(x+1)(4x−2)
⇒ 8x+16=4x
2
−2x+4x−2
⇒ 4x
2
−6x−18=0
⇒ 2x
2
−3x−9=0
⇒ (x−3)(2x+3)=0
∴ x=3 or x=
2
−3
(2) For Fig.(b),
AB∥CD [ Given ]
∴ Quadrilateral ABCD is a trapezium.
∴
CO
AO
=
DO
BO
[ Diagonals of trapezium divides each other proportionally ]
⇒
5x−3
3x−1
=
6x−5
2x+1
⇒ (3x−1)(6x−5)=(2x+1)(5x−3)
⇒ 18x
2
−15x−6x+5=10x
2
−6x+5x−3
⇒ 8x
2
−20x+8=0
⇒ 2x
2
−5x+2=0
⇒ (x−2)(2x−1)=0
∴ x=2 or x=
2
1
(3) For Fig.(c),
AB∥CD [ Given ]
∴ Quadrilateral ABCD is a trapezium.
∴
CO
AO
=
DO
BO
[ Diagonals of trapezium divides each other proportionally ]
⇒
x−3
3x−19
=
4
x−4
⇒ (3x−19)(4)=(x−4)(x−3)
⇒ 12x−76=x
2
−3x−4x+12
⇒ x
2
−19x+88=0
⇒ (x−8)(x−11)=0
∴ x=8 or x=11
Step-by-step explanation:
Answer:
search-icon-header
Search for questions & chapters
search-icon-image
Class 10
>>Maths
>>Triangles
>>Criteria for Triangle Similarity
>>(i) In Fig. (a), if AB ∥ CD, find the va
Question
Bookmark
(i) In Fig. (a), if AB ∥ CD, find the value of x.
(ii) In Fig. (b), if AB ∥ CD, find the value of x.
(iii) If Fig. (c), AB ∥ CD. if OA = 3x - 19, OB = x - 4, OC = x - 3 and OD = 4,find x.
969165
expand
Medium
Solution
verified
Verified by Toppr
(1) For Fig.a,
AB∥CD [ Given ]
∴ Quadrilateral ABCD is a trapezium.
∴
CO
AO
=
DO
BO
[ Diagonals of trapezium divides each other proportionally ]
⇒
4x−2
4
=
2x+4
x+1
⇒ 4(2x+4)=(x+1)(4x−2)
⇒ 8x+16=4x
2
−2x+4x−2
⇒ 4x
2
−6x−18=0
⇒ 2x
2
−3x−9=0
⇒ (x−3)(2x+3)=0
∴ x=3 or x=
2
−3
(2) For Fig.(b),
AB∥CD [ Given ]
∴ Quadrilateral ABCD is a trapezium.
∴
CO
AO
=
DO
BO
[ Diagonals of trapezium divides each other proportionally ]
⇒
5x−3
3x−1
=
6x−5
2x+1
⇒ (3x−1)(6x−5)=(2x+1)(5x−3)
⇒ 18x
2
−15x−6x+5=10x
2
−6x+5x−3
⇒ 8x
2
−20x+8=0
⇒ 2x
2
−5x+2=0
⇒ (x−2)(2x−1)=0
∴ x=2 or x=
2
1
(3) For Fig.(c),
AB∥CD [ Given ]
∴ Quadrilateral ABCD is a trapezium.
∴
CO
AO
=
DO
BO
[ Diagonals of trapezium divides each other proportionally ]
⇒
x−3
3x−19
=
4
x−4
⇒ (3x−19)(4)=(x−4)(x−3)
⇒ 12x−76=x
2
−3x−4x+12
⇒ x
2
−19x+88=0
⇒ (x−8)(x−11)=0
∴ x=8 or x=11