Advertisements
Advertisements
Question
In the adjacent figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE
Solution
In ∆ABC and ∆ADE
∠ACB = ∠AED = 90°
∠A = ∠A ...(common)
∴ ∆ABC ~ ∆ADE ...(By AA similarity)
`"BC"/"DE" = "AB"/"AD" = "AC"/"AE"`
`12/"DE" = 13/3 = 5/"AE"`
In ∆ABC, AB2 = BC2 + AC2
= 122 + 52
= 144 + 25
= 169
AB = `sqrt(169)` = 13
Consider, `13/3 = 5/"AE"`
∴ AE = `(5 xx 3)/13 = 15/13`
AE = `15/13` and DE = `36/13`
Consider, `12/"DE" = 13/3`
DE = `(12 xx 3)/13 = 36/13`
APPEARS IN
RELATED QUESTIONS
E and F are points on the sides PQ and PR, respectively, of a ΔPQR. For the following case, state whether EF || QR:
PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
In the given figure, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.
Proof : In ΔXDE, PQ || DE ...`square`
∴ `"XP"/square = square/"QE"` ...(I) (Basic proportionality theorem)
In ΔXEF, QR || EF ...`square`
∴ `square/square = square/square ..."(II)" square`
∴ `square/square = square/square` ...from (I) and (II)
∴ seg PR || seg DF ...(converse of basic proportionality theorem)
In Δ ABC , MN || BC .
If `"AB"/"AM" = 9/4` , find `("Ar" ("trapezium MBCN"))/("Ar" . (triangle "ABC"))`
The length of a river in a map is 54cm. if lcm on the map represents 12500m on land, find the length of the river.
In ΔABC, D and E are the points on sides AB and AC respectively. Find whether DE || BC, if:
- AB = 9 cm, AD = 4 cm, AE = 6 cm and EC = 7.5 cm.
- AB = 6.3 cm, EC = 11.0 cm, AD = 0.8 cm and AE = 1.6 cm.
In ΔABC, point D divides AB in the ratio 5:7, Find: BC, If DE = 2.5cm
Harmeet is 6 feet tall and casts a shadow of 3 feet long. What is the height of a nearby pole if it casts a shadow of 12 feet long at the same time?
The areas of two similar triangles are 169cm2 and 121cm2 respectively. If one side of the larger triangle is 26cm, find the length of the corresponding side of the smaller triangle.
A man whose eye-level is 2 m above the ground wishes to find the height of a tree. He places a mirror horizontally on the ground 20 m from the tree and finds that if he stands at a point C which is 4 m from the mirror B, he can see the reflection of the top of the tree. How height is the tree?
In fig. BP ⊥ AC, CQ ⊥ AB, A−P−C, and A−Q−B then show that ΔAPB and ΔAQC are similar.
In ΔAPB and ΔAQC
∠APB = [ ]° ......(i)
∠AQC = [ ]° ......(ii)
∠APB ≅ ∠AQC .....[From (i) and (ii)]
∠PAB ≅ ∠QAC .....[______]
ΔAPB ~ ΔAQC .....[______]
