Telangana SCERT Solution Class X (10) Maths Chapter 8 Similar Triangles Exercise 8.3
Exercise 8.3
(Q1) D, E, F are mid points of sides BC, CA, AB of △ABC. Find the ratio of areas of △DEF and △ABC.
=> Given: In △ABC
D, E, F are the mid points of sides BC, CA, AB.
D and E are the mid points of BC and AC.
∴ DE || AB => DE || FA ——- (1)
D and D are the mid points BC and AB
∴ DF || AC => DF || A —— (2)
From (1) and (2)
AFDE is a parallelogram & BDEF is a parallelogram.
In △DEF and △ABC
∠BAC = ∠FDF (∵ opposite angels of a parallelogram)
∠ABC = ∠DEF (∵ Opposite angels of a parallelogram)
By A – A similarity
△DEF ~ △ABC
Area of △DEF/Area of △ABC = DE2/AB2
= (1/2 AB)2/AB2 (DE2 = (1/2 AB)2)
= 1/4 AB2/AB2
= 1/4
∴ Area of DEF : Area of △ABC = 1:4
(Q2) In △ABC, XY || AC and XY divides the triangle into two parts of equal area.
Find AX/XB
=> Given: In △ABC,
XY || AC and Area of △XBC = 1/2 Area of △ABC
=> Area of △BC/Area of XBY = 2/1 ——- (1)
In △ABC and △XBY,
∠B = ∠B (common)
∠A = ∠X (Corresponding angles)
By A-A similarity
△ABC ~ △XBY
Area of △ABC/Area of △XBY = AB2/XB2 —— (2)
From (1) and (2)
AB2/XB2 = 2
=> AB/XB = √2
=> AB/XB = 1 = √2 – 1
=> AB-XB/XB = √2 – 1
AX/XB = √2-1/1
AX:XB = √2 -1:1
(Q3) Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Given: △ABC ~ △DEF
AX and DY are the medians of △ABC and △DEF
To prove: Area of ABC/Area of DEF = AX2/DY2
Proof: Area of △ABC/Area of △DEF – AB2/DE2 ——- (1)
△ABC ~ △DEF
=> AB/DE = BC/EF
=> AB/DE = 2BX/2EY (∵ BC = 2BX) (∵ EF = 2EY)
=> AB/DE = BX/EY = —— (2)
△ABC ~ △DEF
AB/DE = BX/EY and ∠B = ∠E
By S.A.S similarity
△ABX ~ △DEY
∴ BX/EY = AX/DY —— (3)
From (2) and (3),
AB/DE = AX/DY
Squaring on both sides,
AB2/DE2 = AX2/DY2 ——- (4)
From (1) and (2)
Area of △ABC/Area of △DEF = AX2/DY2
(Q4) △ABC ~ DEF. BC = 3cm, EF = 4cm and area of △ABC = 54 cm2, determine the area of △DEF.
Given: △ABC ~ △DEF
BC = 3cm and EF = 4cm
Area of △ABC = 54 cm2
We know that,
The ratio of area of two similar triangles is equal to the ratio of the squares of any two corresponding sides
Area of △ABC/Area of △DEF = BC2/EF2
=> 54/Area of △DEF = (3)2/(4)2
=> 54/Area of △DEF = 9/16
=> Area of △DEF = 54 X 16/9
∴ Area of △DEF = 96 cm2
(Q5) ABC is a triangle and PQ is a straight line meeting AB in and Ac in Q. If AP = 1 cm, BP = 3 cm, AQ = 1.5 cm and CQ = 4.5, prove that area of △APQ = 1/16 (Area of △ABC)
=> Given: In △ABC,
PQ is the straight line meeting AB in P and AC in ‘Q’.
AP = 1cm, BP = 3 cm
AQ = 1.5cm and QC = 4.5 cm
To prove: Area of △APQ = 1/16 (Area of △ABC)
Proof: AP = 1cm & PB = 3cm
=> AB = 1+3 = 4cm
AQ = 1.5cm & QC = 4.5cm
=> AC = 1.5 + 4.5 = 6cm
Consider △APQ and △ABC
AP/AB = 1/4 & AQ/AC = 1.5/6 = 1/4
∴ AP/AB = AQ/AC
By converse of basic proportionality theorem,
PQ || BC
By A A similarity
△APQ ~ △ABC
=> Area of △APQ/Area of △ABC = AP2/AB2
= 12/42
=> ∴ Area of △APQ = 1/16 (Area of △ABC)
(Q6) The areas of two similar triangles are 81 cm2 and 49 cm2 respectively. If the altitude of the bigger triangle is 4.5 cm. Find the corresponding altitude of the smaller triangle.
Give: △ABC = 81 cm2 & Area of △DEF = 49 cm2
Altitude of the △ABC,
AX = 4.5 cm
We know that, the ratio of area of two similar triangles is equal to the ratio of squares of corresponding altitudes
∴ Area of △ABC/Area of △DEF = AX2/DY2
=> 81/49 = (4.5)2/DY2
=> (4.5/DY)2 = 81/49
= 4.5/DY = √81/49
=> 4.5/DY = 9/7
=> DY = 4.5 X 7/9
= DY = 3.5 cm
∴ Altitude of the smaller triangle = 3.5 cm.
Here is your solution of Telangana SCERT Class 10 Math Chapter 8 Similar Triangles Exercise 8.3
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