SSC (English Medium) इयत्ता १० वी - Maharashtra State Board Important Questions

In the figure, PQ ⊥ BC, AD ⊥ BC. To find the ratio of A(ΔPQB) and A(ΔPBC), complete the following activity.

Given: PQ ⊥ BC, AD ⊥ BC

Now, A(ΔPQB)  = `1/2 xx square xx square`

A(ΔPBC)  = `1/2 xx square xx square`

Therefore, 

`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`

= `square/square`

In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR. 

Complete the proof by filling in the boxes.

solution:

In ∆PMQ,

Ray MX is the bisector of ∠PMQ.

∴ `("MP")/("MQ") = square/square` .............(I) [Theorem of angle bisector]

Similarly, in ∆PMR, Ray MY is the bisector of ∠PMR.

∴ `("MP")/("MR") = square/square` .............(II) [Theorem of angle bisector]

But `("MP")/("MQ") = ("MP")/("MR")`  .............(III) [As M is the midpoint of QR.] 

Hence MQ = MR

∴ `("PX")/square = square/("YR")`  .............[From (I), (II) and (III)]

∴ XY || QR   .............[Converse of basic proportionality theorem]

In ΔABC, ray BD bisects ∠ABC, A – D – C, seg DE || side BC, A – E – B, then for showing `("AB")/("BC") = ("AE")/("EB")`, complete the following activity:

Proof :

In ΔABC, ray BD bisects ∠B.

∴ `square/("BC") = ("AD")/("DC")`   ...(I) (`square`)

ΔABC, DE || BC

∴ `(square)/("EB") = ("AD")/("DC")`   ...(II) (`square`)

∴ `("AB")/square = square/("EB")`   ...[from (I) and (II)]

In the following figure, in ΔABC, ∠B = 90°, ∠C = 60°, ∠A = 30°, AC = 18 cm. Find BC.

Adjacent sides of a parallelogram are 11 cm and 17 cm. If the length of one of its diagonal is 26 cm, find the length of the other.

In the following figure, AE = EF = AF = BE = CF = a, AT ⊥ BC. Show that AB = AC =  `sqrt3xxa`

If the sides of a triangle are 6 cm, 8 cm and 10 cm, respectively, then determine whether the triangle is a right angle triangle or not.

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, ∠QPR = 90°, seg PM ⊥ seg QR and Q–M–R, PM = 10, QM = 8, find QR.

For finding AB and BC with the help of information given in the figure, complete following activity.

AB = BC ..........

∴ ∠BAC =

∴ AB = BC = × AC

                 = × `sqrt8`

                 = × `2sqrt2`

                 =

In the given figure, ∠DFE = 90°, FG ⊥ ED, If GD = 8, FG = 12, find (1) EG (2) FD and (3) EF

In the given figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ² = 4PM² – 3PR².

Walls of two buildings on either side of a street are parallel to each other. A ladder 5.8 m long is placed on the street such that its top just reaches the window of a building at the height of 4 m. On turning the ladder over to the other side of the street, its top touches the window of the other building at a height 4.2 m. Find the width of the street.

In ∆ABC, AB = 10, AC = 7, BC = 9, then find the length of the median drawn from point C to side AB.

In ∆ABC, point M is the midpoint of side BC. If, AB² + AC² = 290 cm², AM = 8 cm, find BC.

In the given figure, point T is in the interior of rectangle PQRS, Prove that, TS² + TQ² = TP² + TR² (As shown in the figure, draw seg AB || side SR and A-T-B)

Four alternative answers for the following question are given. Choose the correct alternative and write its alphabet:

Out of the dates given below which date constitutes a Pythagorean triplet?

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.