## Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was 15√2 km. Find their speed per hour.

Distance between Pranali and Prasad after 2 hours = 15√2 kmSince they travel at same speed, they have covered same distance.Construction: Draw a triangle PQR such that PQ = PR = x and QR = 15√2P = 90˚In PQR , PQ2+PR2 = QR2 [Pythagoras theorem]x2+x2 = (15√2)22×2 = 2×225×2 = 225x = 15Distance covered by them is 15 km.Given time […]

## Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Construction:Draw a parallelogram ABCD. Let diagonals AC and BD meet at P. To prove : AC2+BD2 = AB2+BC2+CD2+DA2AB = CD [Opposite sides of parallelogram are equal]BC = DA [Opposite sides of parallelogram are equal]Since diagonals of a parallelogram bisect each other,AP = ½ ACBP = ½ BDP is the midpoint of diagonals AC and BD.InABC , […]

## From the information given in the figure 2.31, prove that PM = PN = √3×a

Proof: In PRMGiven MQ = QR = aQ is the midpoint of MR .PQ is the median.PR2+PM2 = 2PQ2+2QM2 [Apollonius theorem]a2+PM2 = 2a2+2a2PM2 = 3a2PM = √3a…………(i) In PQNGiven NR = QR = aR is the midpoint of QN.PR is the median.PN2+PQ2 = 2PR2+2RN2 [Apollonius theorem]PN2+a2 = 2a2+2a2PN2 = 3a2PN = √3a…………..(ii)From (i) and (ii) PM = PN = √3×aHence proved.

## ABC is an equilateral triangle. Point P is on base BC such that PC = 1/3 BC, if AB = 6 cm find AP.

Given ABC is an equilateral triangle.PC = (1/3) BCPC = (1/3)×6 [BC = 6, side of equilateral triangle]PC = 2Construction:Draw segment AD⊥BC In ADC, C = 60˚D = 90˚CAD = 30˚ADC is a 30˚- 60˚- 90˚triangleAD = (√3/2)×AC [Side opposite to 60˚]AD = (√3/2)×6AD =3√3cmDC = (1/2)BC [AD⊥BC]DC = (1/2)×6 = 3cmDC = DP+PC [D-P-C]3 […]

## In ABC seg AP is a median. If BC = 18, AB2 + AC2 = 260 Find AP.

Given AP is the median.PC = BC/2PC = 18/2 =9AB2+AC2 = 2AP2+2PC2 [Apollonius theorem]260 = 2AP2+2×922AP2 = 260-2×922AP2 = 260-162AP2 = 68/2 = 49Taking square roots on both sidesAP = 7Hence AP = 7 units.

## Find the length of the side and perimeter of an equilateral triangle whose height is √3 cm.

Let BD be height of the triangle.Since ABC is equilateral, BD is a perpendicular bisector.AD = a/2BD = √3 [given height = √3]AB = aApplying Pythagoras theorem in ABDAB2 = AD2+BD2a2 = (a/2)2+(√3)2a2 = (a2/4)+3(3/4)a2 = 3a2 = 4a = 2Hence length of side of equilateral triangle is 2 cm.Perimeter = 3×2 = 6 [Perimeter of equilateral triangle = 3×side]Hence […]

## Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm.

Let PQRS be the rectangle.Let length be PQ = 16 cmArea of rectangle = Length × BreadthArea of rectangle PQRS = PQ×QR192 = 16×QRQR = 192/16 = 12cmNow in PQR , Q = 90˚ [Angles of a rectangle are 90˚]PR2 = PQ2+QR2 [Pythagoras theorem]PR2 = 162+122PR2 = 256+144 = 400PR = 20Hence the diagonal of the rectangle is 20cm […]

## In RST, S = 90°, T = 30°, RT = 12 cm then find RS and ST.

Given S = 90˚ , T = 30˚R = 180-(90+30) = 60˚ [Sum of angles of triangle is equal to 180˚]RST is a 30˚ – 60˚- 90˚ triangleRS = ½ RT [Side opposite to 30˚]RS = ½ ×12 = 6ST = √3/2 RT [side opposite to 60˚]ST = (√3/2 )×12ST = 6√3Hence RS = 6 […]

## Some questions and their alternative answers are given. Select the correct alternative.

(A) (1, 5, 10)Here 12+52 ≠ 102The square of largest number is not equal to sum of squares of the other two numbers.So (1, 5, 10) is not a Pythagorean triplet.(B) (3, 4, 5)Here 32+42 = 52The square of largest number is equal to sum of squares of the other two numbers.So (3, 4, 5) is a […]