find the height of isosceles trapezium in which parallel sides are 24cm and 12cm and non parallel sides are 10cm and 10cm. option are (i)8cm (ii)10cm (iii)6cm (iv)12cm
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GIVEN: An isosceles trapezium ABCD, AD = BC , AB//DC, AB = 10cm, DC = 4 cm.
DM & CN are perpendicular to AB.
=> DM// CN ( lines perpendicular to the same line are // to each other)
=> DCNM is a rectangle,
So, DC = MN = 4 cm
And right tri DMA is congruent to right tri CNB ( by RHS congruency criterion)
=> AM = BN = (10–4)/2 = 3 cm . . . . . . . .(1)
Perimeter of ABCD = 2x + 10 + 4 = 24
=> 2x = 10
=> x = 5 cm or AD = 5cm . . . . . . . . . . . .(2)
In right tri AMD , by (1) & (2)
DM² = 5² - 3² = 25 - 9 = 16 ( by pythagoras law)
=> DM or distance between // sides = √16 = 4 cm
=> Ans: 4 cm
DM & CN are perpendicular to AB.
=> DM// CN ( lines perpendicular to the same line are // to each other)
=> DCNM is a rectangle,
So, DC = MN = 4 cm
And right tri DMA is congruent to right tri CNB ( by RHS congruency criterion)
=> AM = BN = (10–4)/2 = 3 cm . . . . . . . .(1)
Perimeter of ABCD = 2x + 10 + 4 = 24
=> 2x = 10
=> x = 5 cm or AD = 5cm . . . . . . . . . . . .(2)
In right tri AMD , by (1) & (2)
DM² = 5² - 3² = 25 - 9 = 16 ( by pythagoras law)
=> DM or distance between // sides = √16 = 4 cm
=> Ans: 4 cm
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