Advertisements
Advertisements
प्रश्न
In the given figure, angle ACP = ∠BDP = 90°, AC = 12 m, BD = 9 m and PA= PB = 15 m. Find:
(i) CP
(ii) PD
(iii) CD
Advertisements
उत्तर
Given : AC = 12 m
BD = 9 m
PA = PB= 15 m
(i) In right angle triangle ACP
(AP)2 = (AC)2 + (CP)2
152 = 122 + CP2
225 = 144 + CP2
225 – 144 = CP2
81 = CP
`sqrt81` = CP
∴ CP = 9 m
(ii) In right angle triangle BPD
(PB)2 = (BD)2 + (PD)2
(15)2 = (9)2 + PD2
225 = 81 + PD2
225 – 81 = PD2
144 = PD2
`sqrt144` = PD2
∴ PD = 12 m
(iii) CP = 9 m
PD = 12 m
∴ CD = CP + PD
= 9 + 12
= 21 m
APPEARS IN
संबंधित प्रश्न
In triangle ABC, ∠C=90°. Let BC= a, CA= b, AB= c and let 'p' be the length of the perpendicular from 'C' on AB, prove that:
1. cp = ab
2. `1/p^2=1/a^2+1/b^2`
In the given figure, ∆ABC is an obtuse triangle, obtuse-angled at B. If AD ⊥ CB (produced) prove that AC2 = AB2 + BC2 + 2BC · BD.
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
Identify, with reason, if the following is a Pythagorean triplet.
(5, 12, 13)
In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that:
4(BL2 + CM2) = 5 BC2
In an isosceles triangle ABC; AB = AC and D is the point on BC produced.
Prove that: AD2 = AC2 + BD.CD.
ABC is a triangle, right-angled at B. M is a point on BC.
Prove that: AM2 + BC2 = AC2 + BM2
In triangle ABC, ∠B = 90o and D is the mid-point of BC.
Prove that: AC2 = AD2 + 3CD2.
Triangle ABC is right-angled at vertex A. Calculate the length of BC, if AB = 18 cm and AC = 24 cm.
In ΔABC, AD is perpendicular to BC. Prove that: AB2 + CD2 = AC2 + BD2
