20. In the given figure, if $AB = PQ$ and $PQ = XY$, then $AB = XY$. State true or false. Give reason for your answer. Step 1: Analyze the given statements. We are given two equalities: $AB = PQ$ and $PQ = XY$. We need to determine if $AB = XY$ is true or false. Step 2: Apply Euclid's Axiom. Euclid's first axiom states: "Things which are equal to the same thing are equal to one another." In this case, $AB$ is equal to $PQ$, and $XY$ is also equal to $PQ$. Since both $AB$ and $XY$ are equal to the same quantity ($PQ$), they must be equal to each other. Step 3: Conclude the truth value. The statement is True. Reason: This follows from Euclid's first axiom, which states that things which are equal to the same thing are equal to one another. $$\boxed{\text{True}}$$ 21. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal. Proof: Step 1: Set up the scenario. Consider two congruent circles with centers $O$ and $O'$. Let their radii be $r$. Let $AB$ be a chord of the first circle and $CD$ be a chord of the second circle. Given that the chords subtend equal angles at their centers, so $\angle AOB = \angle CO'D$. Step 2: Consider the triangles formed by the radii and the chords. In $\triangle AOB$ and $\triangle CO'D$: • $OA = OB = r$ (radii of the first circle) • $O'C = O'D = r$ (radii of the second circle) Since the circles are congruent, their radii are equal, so $OA = OB = O'C = O'D$. Step 3: Apply the Side-Angle-Side (SAS) congruence criterion. We have: • $OA = O'C$ (radii of congruent circles) • $\angle AOB = \angle CO'D$ (given) • $OB = O'D$ (radii of congruent circles) By the SAS congruence criterion, $\triangle AOB \cong \triangle CO'D$. Step 4: Conclude the equality of chords. Since the triangles are congruent, their corresponding parts are equal (CPCTC). Therefore, $AB = CD$. This proves that if chords of congruent circles subtend equal angles at their centers, then the chords are equal. 22. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP = BQ$. Show that: (i) $\triangle APD \cong \triangle CQB$ (ii) $AP = CQ$ Proof: (i) To show $\triangle APD \cong \triangle CQB$: Step 1: Identify properties of a parallelogram. In parallelogram ABCD, $AD \parallel BC$ and $AD = BC$ (opposite sides are parallel and equal). Step 2: Consider $\triangle APD$ and $\triangle CQB$. • $AD = CB$ (Opposite sides of parallelogram ABCD) • $DP = BQ$ (Given) • Since $AD \parallel BC$ and $BD$ is a transversal, the alternate interior angles are equal: $\angle ADB = \angle CBQ$. This means $\angle ADP = \angle CBQ$. Step 3: Apply the Side-Angle-Side (SAS) congruence criterion. We have: • Side $AD = CB$ • Angle $\angle ADP = \angle CBQ$ • Side $DP = BQ$ Therefore, by SAS congruence criterion, $\boxed{\triangle APD \cong \triangle CQB}$. (ii) To show $AP = CQ$: Step 1: Use the result from part (i). We have already proven that $\triangle APD \cong \triangle CQB$. Step 2: Apply the property of Corresponding Parts of Congruent Triangles (CPCTC). Since the triangles $\triangle APD$ and $\triangle CQB$ are congruent, their corresponding sides are equal. Therefore, $\boxed{AP = CQ}$. 23. The following table shows the daily production of toys in a factory. Draw a histogram for the following data. | No. of toys | No. of days | |-------------|-------------| | 0-50 | 5 | | 50-100 | 8 | | 100-150 | 12 | | 150-200 | 5 | To draw a histogram: Step 1: Set up the axes. • Draw a horizontal axis (x-axis) and label it "No. of toys". Mark the class limits: 0, 50, 100, 150, 200. • Draw a vertical axis (y-axis) and label it "No. of days" (frequency). Choose an appropriate scale for the frequency, for example, 1 unit = 1 day. Step 2: Draw the bars. • For the class interval 0-50, draw a rectangular bar with width corresponding to the interval and height corresponding to the frequency 5. • For the class interval 50-100, draw a rectangular bar with width corresponding to the interval and height corresponding to the frequency 8. This bar should be adjacent to the previous one. • For the class interval 100-150, draw a rectangular bar with width corresponding to the interval and height corresponding to the frequency 12. This bar should be adjacent to the previous one. • For the class interval 150-200, draw a rectangular bar with width corresponding to the interval and height corresponding to the frequency 5. This bar should be adjacent to the previous one. The bars in a histogram are always adjacent to each other, as the class intervals are continuous. 24. Write any four solutions for the linear equation $3x + y - 12 = 0$. Step 1: Rewrite the equation to express one variable in terms of the other. It's easier to express $y$ in terms of $x$: $$y = 12 - 3x$$ Step 2: Choose arbitrary values for $x$ and calculate the corresponding $y$ values. • Solution 1: Let $x = 0$. $$y = 12 - 3(0)$$ $$y = 12 - 0$$ $$y = 12$$ So, $(0, 12)$ is a solution. • Solution 2: Let $x = 1$. $$y = 12 - 3(1)$$ $$y = 12 - 3$$ $$y = 9$$ So, $(1, 9)$ is a solution. • Solution 3: Let $x = 2$. $$y = 12 - 3(2)$$ $$y = 12 - 6$$ $$y = 6$$ So, $(2, 6)$ is a solution. • Solution 4: Let $x = 4$. $$y = 12 - 3(4)$$ $$y = 12 - 12$$ $$y = 0$$ So, $(4, 0)$ is a solution. Four solutions for the equation $3x + y - 12 = 0$ are: $$\boxed{(0, 12), (1, 9), (2, 6), (4, 0)}$$