SSC (Marathi Semi-English) 10th Standard Board Exam [इयत्ता १० वी] - Maharashtra State Board Important Questions for Geometry Mathematics 2

In ΔABC, B-D-C and BD = 7, BC = 20, then find the following ratio.

(i) `(A(ΔABD))/(A(ΔADC))`

(ii) `(A(ΔABD))/(A(ΔABC))`

(iii) `(A(ΔADC))/(A(ΔABC))`

Prove that, The areas of two triangles with the same height are in proportion to their corresponding bases. To prove this theorem start as follows:

1. Draw two triangles, give the names of all points, and show heights.
2. Write 'Given' and 'To prove' from the figure drawn.

From the information given in the figure, determine whether MP is the bisector of ∠KMN.

If ΔABC ∼ ΔDEF such that ∠A = 92° and ∠B = 40°, then ∠F = ?

If ΔABC ∼ ΔDEF, length of side AB is 9 cm and length of side DE is 12 cm, then find the ratio of their corresponding areas.

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`

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

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

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.

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.