Advertisements
Advertisements
प्रश्न
In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 − OD2 − OE2 − OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
उत्तर
(i) Applying Pythagoras theorem in ΔAOF, we obtain
OA2 = OF2 + AF2
Similarly, in ΔBOD
OB2 = OD2 + BD2
Similarly, in ΔCOE
OC2 = OE2 + EC2
Adding these equations,
OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE2 + EC2
OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2.
From above result
AF2 + BD2 + EC2 = (OA2 - OE2) + (OC2 - OD2) + (OB2 - OF2)
∴ AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
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`
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM . MR
Identify, with reason, if the following is a Pythagorean triplet.
(4, 9, 12)
In the figure: ∠PSQ = 90o, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.
In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
Prove that: 2AC2 - AB2 = BC2 + CD2 + DA2
A man goes 10 m due east and then 24 m due north. Find the distance from the straight point.
In an equilateral triangle ABC, the side BC is trisected at D. Prove that 9 AD2 = 7 AB2.
A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?
If ΔABC ~ ΔPQR, `("ar" triangle "ABC")/("ar" triangle "PQR") = 9/4` and AB = 18 cm, then the length of PQ is ______.
In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC2 + BD2 = AD2 + BC2
[Hint: Produce AB and DC to meet at E.]
