\(\square\)ABCD is a rectangle. Taking AD as a diameter, a semicircle AXD is drawn which intersects the diagonal BD at X. If AB = 12 cm, AD = 9 cm, then find the values of BD and BX.
Show Hint
Solution and Explanation
Step 1: Understanding the Concept:
We can use the Pythagorean theorem in the right-angled triangle formed by the sides of the rectangle to find the length of the diagonal BD. Then, using the property that the angle in a semicircle is a right angle, we can find BX.
Step 2: Key Formula or Approach:
1. Pythagorean Theorem: In a right-angled triangle, \((\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Height})^2\).
2. Angle in a Semicircle Theorem: The angle subtended by a diameter at any point on the circumference is a right angle (90\(^\circ\)).
3. Property of Right-Angled Triangles: In a right-angled triangle, if an altitude is drawn to the hypotenuse, then \((\text{Leg})^2 = (\text{Adjacent segment of hypotenuse}) \times (\text{Whole hypotenuse})\).
Step 3: Detailed Explanation:
Given: \[\begin{array}{rl} \bullet & \text{ABCD is a rectangle, so \(\angle DAB = 90^\circ\).} \\ \bullet & \text{AB = 12 cm, AD = 9 cm.} \\ \end{array}\] 1. Find the value of BD: In right-angled \(\triangle\)DAB, BD is the hypotenuse. By Pythagorean theorem: \[ BD^2 = AB^2 + AD^2 \] \[ BD^2 = 12^2 + 9^2 = 144 + 81 = 225 \] \[ BD = \sqrt{225} = 15 \text{ cm} \] 2. Find the value of BX: A semicircle is drawn with AD as the diameter. The diagonal BD intersects the semicircle at point X. According to the Angle in a Semicircle Theorem, the angle subtended by the diameter AD at point X on the circumference is 90\(^\circ\). \[ \therefore \angle AXD = 90^\circ \] This means that seg AX is perpendicular to the diagonal BD. Now, consider the right-angled \(\triangle\)DAB. seg AX is the altitude from vertex A to the hypotenuse BD. Using the property of right-angled triangles (related to geometric mean): \[ AB^2 = BX \times BD \] We have AB = 12 cm and we found BD = 15 cm. \[ 12^2 = BX \times 15 \] \[ 144 = 15 \times BX \] \[ BX = \frac{144}{15} = \frac{48}{5} = 9.6 \text{ cm} \]
Step 4: Final Answer:
The length of BD is 15 cm and the length of BX is 9.6 cm.
Top Questions on Circles
- Let PQ and MN be two straight lines touching the circle \(x^2+y^2-4x-6y-3=0\) at the points A and B respectively. Let O be the centre of the circle and \(\angle AOB = \pi/3\). Then the locus of the point of intersection of the lines PQ and MN is:
- Let the circle \(x^2+y^2=4\) intersect the \(x\)-axis at points \(A(a,0)\) and \(B(b,0)\). Let \(P(2\cos\alpha,2\sin\alpha)\), \(0<\alpha<\frac{\pi}{2}\), and \(Q(2\cos\beta,2\sin\beta)\) be two points on the circle such that \((\alpha-\beta)=\frac{\pi}{2}\). Then the point of intersection of lines \(AQ\) and \(BP\) lies on:
- If the sum of the diameters of two circles is 40 cm and the difference of their radii is 6 cm, then the ratio of the area of the smaller circle to that of the bigger circle is:
- The absolute difference between the squares of the radii of the two circles passing through the point \( (-9, 4) \) and touching the lines \( x + y = 3 \) and \( x - y = 3 \), is equal to:
- Let $ ABC $ be the triangle such that the equations of lines $ AB $ and $ AC $ are: $ 3y - x = 2 \quad \text{and} \quad x + y = 2, $ respectively, and the points $ B $ and $ C $ lie on the x-axis. If $ P $ is the orthocentre of the triangle $ ABC $, then the area of the triangle $ PBC $ is equal to:
Questions Asked in Maharashtra Class X Board exam
- Appreciation of the poem:
Read the following poem and write an appreciation of it with the help of the points given below:
Stopping by Woods on a Snowy Evening
Whose woods these are I think I know.
His house is in the village, though;
He will not see me stopping here
To watch his woods fill up with snow
My little horse must think it queer
To stop without a farmhouse near
Between the woods and frozen lake
The darkest evening of the year.
He gives his harness bells a shake
To ask if there is some mistake.
The only other sound’s the sweep
Of easy wind and downy flake.
The woods are lovely, dark and deep,
But I have promises to keep,
And miles to go before I sleep,
And miles to go before I sleep.
Robert Frost
- Maharashtra Class X Board - 2025
- Reading Comprehension
- Pick out the infinitive from the following sentence :
He was asking to go for the concert.- Maharashtra Class X Board - 2025
- Grammar and Language Usage
- Out of the following which is a Pythagorean triplet ?
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\).
- Maharashtra Class X Board - 2025
- Properties of Triangles
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\))
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.