Advertisements
Advertisements
प्रश्न
In ∆ABC, point M is the midpoint of side BC. If, AB2 + AC2 = 290 cm2, AM = 8 cm, find BC.
Advertisements
उत्तर
In ∆ABC, point M is the midpoint of side BC.
AB2 + AC2 = 2AM2 + 2BM2 ...[by Apollonius theorem]
290 = 2 (8)2 + 2BM2
290 = 2 (64) + 2BM2
290 = 128 + 2BM2
2BM2 = 290 - 128
2BM2 = 162
BM2 = 81
BM = `sqrt81`
BM = 9 cm
Point M is the midpoint of BC
∴ BC = 2BM = 2 × 9
∴ BC = 18 cm.
संबंधित प्रश्न
In the given figure, seg PS is the median of ∆PQR and PT ⊥ QR. Prove that,
PR2 = PS2 + QR × ST + `("QR"/2)^2`
Out of the following, which is the Pythagorean triplet?
Find the perimeter of a square if its diagonal is `10sqrt2` cm:
Some question and their alternative answer are given. Select the correct alternative.
Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude.
Some question and their alternative answer are given. Select the correct alternative.
In ∆ABC, AB = \[6\sqrt{3}\] cm, AC = 12 cm, BC = 6 cm. Find measure of ∠A.
Find the length a diagonal of a rectangle having sides 11 cm and 60 cm.
A side of an isosceles right angled triangle is x. Find its hypotenuse.
In ∆RST, ∠S = 90°, ∠T = 30°, RT = 12 cm, then find RS and ST.
Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM
If hypotenuse of a right angled triangle is 5 cm, find the radius of
the circle passing through all vertices of the triangle.
Choose the correct alternative:
Out of the following which is a Pythagorean triplet?
In ΔPQR, seg PM is a median, PM = 9 and PQ2 + PR2 = 290. Find the length of QR.
In ΔPQR, seg PM is the median. If PM = 9, PQ2 + PR2 = 290, Find QR.
Out of given triplets, which is not a Pythagoras triplet?
Out of all numbers from given dates, which is a Pythagoras triplet?
Which of the following figure is formed by joining the mid-points of the adjacent sides of a square?
In the given figure, triangle ABC is a right-angled at B. D is the mid-point of side BC. Prove that AC2 = 4AD2 – 3AB2.
