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

Prove that “That ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.”

Write down the equation of a line whose slope is 3/2 and which passes through point P, where P divides the line segment AB joining A(-2, 6) and B(3, -4) in the ratio 2 : 3.

ΔRST ~ ΔUAY, In ΔRST, RS = 6 cm, ∠S = 50°, ST = 7.5 cm. The corresponding sides of ΔRST and ΔUAY are in the ratio 5 : 4. Construct ΔUAY.

Construct the circumcircle and incircle of an equilateral ∆XYZ with side 6.5 cm and centre O. Find the ratio of the radii of incircle and circumcircle.

∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR, such that `"PQ"/"LT" = 3/4`.

Find distance between point Q(3, – 7) and point R(3, 3)

Solution: Suppose Q(x₁, y₁) and point R(x₂, y₂)

x₁ = 3, y₁ = – 7 and x₂ = 3, y₂ = 3

Using distance formula,

d(Q, R) = `sqrt(square)`

∴ d(Q, R) = `sqrt(square - 100)`

∴ d(Q, R) =  `sqrt(square)`

∴ d(Q, R) = `square`

Prove that sin⁶θ + cos⁶θ = 1 – 3 sin²θ. cos²θ.

Prove that:

sec²θ + cosec²θ = sec²θ x cosec²θ

Prove that: If the angles of a triangle are 45° – 45° – 90°^, then each of the perpendicular sides is \[\frac{1}{\sqrt{2}}\]times the hypotenuse.”

Eliminate θ, if
x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ

Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`

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

sin θ × cosec θ = ______

Find the circumferences of a circle whose radius is 7 cm.

The radii of the ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its curved surface area.

The radius of a circle is 10 cm. The measure of an arc of the circle is 54^°. Find the area of the sector associated with the arc. (\[\pi\]= 3.14 )

In the given figure, if A(P-ABC) = 154 cm²  radius of the circle is 14 cm, find

(1) `∠APC` 

(2) l ( arc ABC) .

In the given figure, square ABCD is inscribed in the sector A - PCQ. The radius of sector C - BXD is 20 cm. Complete the following activity to find the area of shaded region 

The ratio of the areas of two triangles with the common base is 14 : 9. Height of the larger triangle is 7 cm, then find the corresponding height of the smaller triangle.

The ratio of the areas of two triangles with common base is 6:5. Height of the larger triangle of 9 cm, then find the corresponding height of the smaller triangle.

Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.