Advertisements
Advertisements
प्रश्न
In ΔPQR, seg PM is a median, PM = 9 and PQ2 + PR2 = 290. Find the length of QR.
Seg PM is a median of ΔPQR, PM = 9 and PQ2 + PR2 = 290, then find QR.
उत्तर
Given: PM = 9,
PQ2 + PR2 = 290
To find QR
∵ PM is the median of QR.
QM = MR ...(i)
So, by apollonius theorem,
PQ2 + PR2 = 2 PM2 + 2 QM2
290 = 2(9)2 + 2(QM)2
290 = 2[81 + (QM)2]
145 = 81 + QM2
QM2 = 145 – 81
QM2 = 64
QM = 8
QM = MR = 8 units ...[From (i)]
QR = QM + MR ...[∵ QNR is a straight line]
QR = 8 + 8
QR = 16 units.
संबंधित प्रश्न
In ∆ABC, point M is the midpoint of side BC. If, AB2 + AC2 = 290 cm2, AM = 8 cm, find BC.
Out of the following, which is the Pythagorean triplet?
Some question and their alternative answer are given. Select the correct alternative.
Find perimeter of a square if its diagonal is \[10\sqrt{2}\]
Find the height of an equilateral triangle having side 2a.
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.
Sum of the squares of adjacent sides of a parallelogram is 130 sq.cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal.
Seg PM is a median of ∆PQR. If PQ = 40, PR = 42 and PM = 29, find QR.
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.
In ΔPQR, seg PM is the median. If PM = 9, PQ2 + PR2 = 290, Find QR.
Choose the correct alternative:
Out of given triplets, which is not a Pythagoras triplet?
"The diagonals bisect each other at right angles." In which of the following quadrilaterals is the given property observed?
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.
