CBSE Class 10 Mathematics Solved Question Paper 2017

The angle of elevation of the sun from the ground is θ.
We have, x:y =√3:1
Now, In ∆ABC

Given: PA and PB are the tangents to the circle.
PA = 12 cm
QC = QD = 3 cm

To find: PC + PD

PA = PB = 12 cm
(The lengths of tangents drawn from an external point to a circle are equal)
Similarly, QC = AC = 3 cm
and QD = BD = 3 cm.

Now, PC = PA − AC = 12 − 3 = 9 cm
Similarly, PD = PB − BD = 12 − 3 = 9 cm

Hence, PC + PD = 9 + 9 = 18 cm.

Let the radius of the hemisphere be x units.
Volume of a hemisphere = Surface area of the hemisphere

Hence, the diameter of the hemisphere is equal to 9 units.

Let the points X (4,k) and Y(1,0)
It is given that the distance XY is 5 units.
By using the distance formula,

S = {−3, −2, −1, 0, 1, 2, 3}
Let E be the event of getting a number whose square is less than or equal to 1.
So, E = {−1, 1, 0}
P(E)=3/7.
Hence, the probability of getting a number whose square is less than or equal to is 3/7.

1) Draw a line segment AB = 8 cm.
2) Draw a ray AX making an acute angle ∠BAX=60° with AB.
3) Draw a ray BY parallel to AX by making an acute angle ∠ABY=∠BAX.
4) Mark four points A₁, A₂, A₃, A₄ on AX and five points B₁,B₂,B₃,B₄,B₅ on BY in such a way that AA₁=A₁A₂=A₂A₃=A₃A₄.
5) Join A₄B₅
6) Let this line intersect AB at a point P.
Thus, P is the point dividing the line segment AB internally in the ratio of 4:5.

The given equation is x²+k(2x+k−1)+2=0.
⇒x²+2kx+k(k−1)+2=0
So, a = 1, b = 2k, c = k(k − 1) + 2
We know D=b²−4ac
⇒D=(2k)²−4×1×[k(k−1)+2]
⇒D=4k²−4[k2−k+2]
⇒D=4k²−4k2+4k−8
⇒D=4k−8=4(k−2)
For equal roots, D = 0
Thus, 4(k − 2) = 0
So, k = 2.

The first term between 200 and 500 divisible by 8 is 208, and the last term is 496.
So, first term (a) = 208
Common difference (d) = 8
a_n=a+(n−1)d=496
⇒208+(n−1)8=496
⇒(n−1)8=288
⇒n−1=36⇒n=37
Hence, there are 37 integers between 200 and 500 which are divisible by 8.

Let AB be the diameter of a circle, with centre O. The tangents PQ and RS are drawn at points A and B, respectively.

We know that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ OA ⊥ PQ and OB ⊥ RS
⇒ ∠OBR = 90°
∠OBS = 90°
∠OAP = 90°
∠OAQ = 90°
We can observe the following:
∠OBR = ∠OAQ and ∠OBS = ∠OAP
Also, these are the pair of alternate interior angles.
Since alternate interior angles are equal, the lines PQ and RS are parallel to each other.
Hence, proved.